equilibrium review Flashcards
intro to equilibrium
Chemical reactions occur via collisions of reactants; these collisions must have sufficient energy to break the bond holding the molecule together and overcome the activation energy barrier. This leads to the formation of products.
Note as the concentration of the reactant decreases over time, the concentration of products increases. With enough energy to overcome the Ea barrier and correct orientation, there are reactants that are reformed as well.
When both forward rate = reverse rate we have reached equilibrium (the plateau stage in the graph)
Consider the rates of reaction at different times during the reaction. The rate is faster at the beginning but slows down as more and more products are made.
So far, all the reactions proceed to completion that is why the single headed arrow is used. But this is not necessarily true for every reaction, many reactions stop far from completion and
3 categories:
- Reactions that favour reactants very strongly. Mixing reactants has no observable results. Example: 2CaO(s) –> 2Ca(s) + O2(g) @25°C. formation of CaO is favoured more than the products so usually just a single headed arrow to CaO
- Reactions that favour products very strongly. Essentially this reaction is considered to go to completion (quantitative reaction). A single arrow is used to indicate that the effect of the reverse reaction is negligible. Example: 2H2(g) + O2(g) –> 2H2O(g)
- Reactions that achieve noticeable equilibrium condition. Both the reactant and product species are always present in a mixture in a closed system. Example: N2O4(g) –> 2NO2(g). both of these are present and being formed
Why is the equilibrium position of a reaction towards: left, right or middle? Many factors affect this outcome but a chemical system will seek the lowest possible energy. This is at equilibrium, because at equilibrium there is no driving force to go either way in reaction direction.
CHARACTERISTICS OF AN EQUILIBRIUM:
- Closed system – no matter can enter or leave the system.
- Dynamic – reactions are occurring in both directions. Constant formation of products and dissociation of products to reactants.
- Reversibility – regardless of whether equilibrium is approached from either pure products or pure reactants, in an equilibrium mixture the compositions (reactant and products will be present). will always reach equilibrium no matter what it starts with
- Reaction rates are equal – the rates of forward reaction (formation of products) equal the rate of reverse reactions (formation of reactants).
R <–> P
Therefore rf = rr
But [R] does not equal [P] (depends on stoich)
However, concentrations of [R] and [P] are constant
Homogeneous equilibrium – all reacting species are in the same phase.
Example: 2H2(g) + O2(g) <–> 2H2O(g)
Heterogeneous equilibrium – products and reactants are not in the same phase.
Example: 2CaO(s) <–> 2Ca(s) + O2(g)
phase equilibrium –> same compounds but different phases reaching equillirbium like liquid and gas h2o
A system at equilibrium exhibits no changes in macroscopic properties (cannot observe any changes) but there is a constant change at the microscopic level. Therefore, equilibrium is a DYNAMIC process and is NOT static.
ex. Nitrogen dioxide; NO2 (red-brown gas) exists in equilibrium with dinitrogen tetroxide; N2O4 (clear, colourless gas).
N2O4 + heat –> NO2
Control: at room temperature the equilibrium tube is light red brown in colour.
Ice bath: at 0ºC, the tube is uniformly light yellow.
[N2O4 ]»_space;»[NO2]
Water bath: at room temperature, the tube is the same colour density at control
N2O4 + heat ⇄ NO2
Hot water bath: at approximately 85ºC, the tube is a dark red orange color.
[N2O4 ] «<[NO2]
We can now add the concept of enthalpy to equilibrium, temperature will shift the equilibrium position.
Conclusions: Equilibrium is dynamic (it is reversible), it responds to changes in temperature.by changing the concentrations of reactants or products. The changes in ratios of reactants or products lead to changes in colour of the equilibrium mixture. Therefore, there is a constant reaction between reactants and products
Law of Mass Action
Determining the Equilibrium constant and Reaction kinetics
A + B C + D
Therefore, the rate law equations for both forward and reverse reactions are:
rf = kf [A][B]
rr = kr [C][D]
At equilibrium the rf = rr, then kf [A][B] = kr [C][D]
kf/kr = [c][d]/[a][b]
Keq = [products]/[reactants]
Big “K” or the equilibrium constant is dependent on temperature ONLY.
Activation energy and equilibrium constant:
The activation energy (Ea) of the forward and reverse reaction affects the magnitude of the product produced.
- Case 1: If the activation energy of the forward reaction is small, much product exists at equilibrium and the “K” constant is large. since k = [product] increased/ [reatcnat] decreased so k is large value. (formation is easy to products when exo since less Ea so more products formed than reactants)
- Case 2: If the activation energy of the reverse reaction is small, much reactant exists at equilibrium and the “K” constant is small. since prpdct decrease/ reactant increase, k is small value. (formation is easy to reactants from products in endo reactant so there is more reactants than products)
- Case 3: If the activation energy of the reverse and forward reaction is the same, then “K” is equal to 1. since similar values divided by each other.
Equilibrium Law or Law of Mass Action:
- Norwegian chemists: Cato Maximilian Guldberg and Peter Waage proposed the law of mass action in 1864 as a general description of the equilibrium condition.
- Law of Mass Action: In a chemical system at equilibrium, there is a constant ratio between the concentrations of the products and the concentrations of the reactants as long as the temperature doesn’t change.
Given the following chemical equation:
jA + kB –> lC + mD
The following can represent the Law of Mass Action or Equilibrium Law is:
K = [c]^l[d]^m/[a]^j[b]^k
Where:
[ ] : Concentration of compound in mol/L
K: Equilibrium constant
j, k, l, m: (coefficients) = exponents of equilibrium law expression
A, B: reactants
C, D: products
What physical states of the substances are included in the Law of Mass Action?
- compounds in gas or aqueous phase
- (species in highest state of entropy or exclude pure substances (liquids or solids))
- pure substances like liquid water or a solid substance are excluded in the law of mass action equation since their concentration doesn’t change when you add more or less of it.
Example: CO(g) + 2H2(g) <–> CH3OH(g)
law of mass action for forward reaction is Keq = [ch3oh]1/[co]1[h2]2 and the reverse reaction is inverse/flipped of this equation.
The constant K can be calculated for a SPECIFIC TEMPERATURE by using equilibrium concentrations of reaction. if not specified of which way reaction to find the mass of action, assume forward reaction (reactants to products). plug in the equilibrium values given and apply exponents of coefs in the balanced equation to find K. K doesn’t change unless temp is changed. equilibrium reactions are reversible. regardless of []’s, k remains the same at the same temp.
** It doesn’t matter if we start with just pure reactants or pure products or even a mixture, equilibrium always produces the same ratio of products to reactants or the same K value.
Chemical Equilibrium:
It is important to remember that equilibrium can be physical or chemical, that is, there can be equilibrium between two different phases (vapour and liquid) or between two compounds in a chemical reaction.
In chemical reactions, it was possible to find information like concentration by using stoichiometry.
However, reactions that are reversible like equilibrium require a different approach. In the case of equilibrium, the ICE table is used.
I: Initial
C: Change - (x)
E: Equilibrium - ([eq’m] only)
- use the given values, place in ice chart. change relating to x values is dependent on coef values in the balanced equation and is negative or adding based on what is being produced (pos if produced). initial + change gives the equilibrium.
- Changes in equilibrium concentration is not known, thus using x with the coefficients of balanced equation will serve as the change.
- balance the equation, create ice table and x * stoich coef is change. plug in/rearrange to find needed value, usually x. you can then also find the equollirbium constant using the equilibrium concentrations found and using the coef as the exponents.
Conclusions about equilibrium expressions:
jA + kB lC + mD
Equilibrium expression:
K = [c]^l[d]^m/[a]^j[b]^k
Reverse reaction, the new equilibrium expression is:
K1 = [a]^j[b]^k/ [c]^l[d]^m
If the original reaction is multiplied with some factor ‘n’ to give:
njA + nkB nlC + nmD
Then, the equilibrium expression becomes:
K = [c]^nl[d]^nm/[a]^nj[b]^nk
Units of the K are determined by the powers of the various terms within the equilibrium expression. K for each unit will depend on the reaction being considered.
Note:
K is temperature dependent
K does NOT provide information about rate (just says what side is favoured and the ratio of reactants to products at equillirbium)
K relates concentrations of reactants or products at equilibrium
Heterogenous equilibria
Homogeneous equilibria: means all reactants and products are in the same phase.
Heterogeneous equilibria: means reactants and products are in different phases.
Example: Electrolysis of water 2H2O(l) <==> 2H2(g) + O2(g)
The equilibrium expression would be:
k = [H2]^2[o2]^1/[H2O]^2 since h2o is pure substance 1g/1ml and c = n/v then (1000g/18.02g/mol) / 1000ml = 55.5 M which is constant concentration
However, the concentration of a pure liquid such as water cannot change. It is fixed and equal to the substance (water) density. For example at SATP, water has a density of 1.00 kg/L. This density when converted into moles/litres gives 55.5 mol/l. Thus adding water or removing water will not change its concentration, therefore it is a constant and is incorporated into the equilibrium constant. Thus, the expression can be re-written as:
K = [H2]^2[o2]^1
only gasses and aqueous are included
in a double displacement reaction where there is aqueous and solid write the net ionic equation, cancelling out whatever is same, including their states on both sides and include only the aqueous from the net ionic equation in the equillirbium expression.
Conclusion: constant concentrations of liquids, solids, as well as spectator ions do not appear in the equilibrium expression.
Percent Reaction
Percent reaction – the yield of product measured at equilibrium compared with the maximum possible yield of product.
(how much reactant is formed to products)
Description of equilibrium - Position of equilibrium
No rxn - <1%
Reactants favoured - <50%
Products favoured - >50%
Quantitative (goes to completion) - >99%
The state of equilibrium can be described in two ways:
- Percentage reaction
- Equilibrium constants
Percent reaction for a product only can be calculated using values from the actual product concentration at equilibrium and theoretical product concentration using stoichiometry.
Percent reaction = actual product yield / theoretical product yield * 100% or [products] eq’m / [products] stoich * 100%
- find stoich value and ice table value of the product equilibrium and concentration. then plug into equation to see. the number is percentage of reactants forming into products and see what is favoured. can check by plugging in eq’m values into equilibrium law
Equilibrium Constant (K) versus the Reaction Quotient (Q)
The equilibrium constant is temperature specific and can only be calculated for values of a chemical reaction AT equilibrium
The reaction quotient (Q) can be calculated using values of a chemical reaction at any given time (initial, final, equilibrium).
Q = [products]/[reactants] at any condition so similar to K except K needs to be at eq’m
There is a relationship that exists between Q and K
- When Q = K, the system is at equilibrium.
- When Q > K, the system must shift left (towards the reactants) to reach equilibrium.
- When Q < K, the system must shift right (towards the products) to reach equilibrium.
for Q use any values to expo of the coefs and solve and determine what relationship it has to K. (if Q = K then at equilibrium, if Q if greater then shift left if Q if smaller then shift right)
Equilibrium Calculations
Steps to a successful equilibrium calculation:
- Always start with a balanced chemical equation.
- Setup an ICE table.
- Ensure that all information in the ICE table is expressed as concentration (mol/L). (c=n/v)
- Use the coefficients in the balanced chemical equation to show the change.
- Solve for x in the equilibrium law expression for the reaction.
- Calculate the equilibrium values by substituting for the value of x.
- Check by substituting calculated equilibrium into the reaction quotient equation (Q). If the value for Q = K then you have done the question properly. If not and check for mathematical errors or if you made an invalid assumption.
(normally calculating for eq’m [] or K)
There are 3 ways of solving an equilibrium question:
- Perfect Squares: If the setup for the equilibrium law is such that it is easy to solve for x by taking the square root or cubed root of both sides, then should be done. However, you can choose to solve for x by using the quadratic equation.
- 500 times Rule: Calculations that involve imperfect squares. In this situation, it is impossible to solve for x by taking the square root of both sides. However, chemists always look for the easy way out. It is possible to make an assumption so long as it is valid to simplify the equation. Comparing the concentration value of a compound to the equilibrium constant can only do this. If there is a difference of 500x between the K value and the concentration value then the assumption becomes valid. Basically, this assumption states that the change in value of the initial versus the equilibrium value is so small that it can be ignored. (only if k is super small you can ignore the 2x etc. check by doing the value/k and if greater than 500 you can ignore)
- Calculations involving quadratic equations. This is the worst case scenario! In this case, one must fully expand the and solve for x using the quadratic equation. The reason this is so is because an assumption CANNOT be made, thus one must solve manually. –> ax2 + bx + c = 0 then sub into -b+-sqare root of b2 - 4ac/2a
Summary of K and Q
- equillibrium is dynamic –> reversing of products to reactants and vice versa
- rf = rr
- Equations have to be balanced
- look at ratio of products to reactants
- k = [products]/[reactants] each to the power of whatever the coefficients of the substance is in the equation
- K is equillibroum constant while small k is rate constant. both are not related
- reverse reaction Keq is just the inverse while multiplying the equation by a factor you have to multiply each power to the same multiple and K (ex *2 then K ^2 or if *1/2 then K ^1/2)
- since K = [p]/[r] if the K is large,greater than one then the product is high and reactnat is small so the eq’m favours products
- if K is small then less products and more reactants so the K is a fraction and eq’m favours reactants
- only [] at eq’m can be inserted into the K expression
- in grade 11 when given molarity, to find molarity of products it was ratios but in grade 12 we set up ice table since its an equilibrium reaction
- ICE –> i is initial concentrations, c is the change by ?x and ? depends on the coef, and E is the equilibrium concentrations
- in all eq’m questions you either solve for x or for k
- Q is same as K expression except the concentration values can be from any condition, not just equilibrium.
- if q = k then at equilibrium
- if Q > K then reaction needs to shift to the left for more reactnats to reach equilibrium
- if Q< K then reaction needs to shift to right for more products to reach eq’m
- when solving for the x values in K expression either see if perfect squares all around, see if you can use rule 500 times which is used when k is very small and seeing if x is small enough to be ignored –> do the initial concentration/k and if greater than 500 valid assumption. Otherwise use quadratic equation
Le Chatelier’s principle
In 1884, French chemist Henry Louis Le Chatelier made observations of chemical systems at equilibrium. He notices that chemical reaction responds to changes made in order to re-establish equilibrium.
Le Chatelier’s Principle – when a chemical system at equilibrium is disturbed by a change in a property, the system adjusts in a way that opposes the change.
Le Chatelier’s Principle and concentration
* Adding more of a reactant to a system at equilibrium and the system will undergo an equilibrium shift (in the opposite direction to make more product).
- increasing a reactnat results in shift right to make more products
- dcreasing a product results on shift right to replenish the product
- increase in a specific substance indicates increase in the product or reactants
- decrease in all [] indicates a volume increase
- simple difference indicates a temperature change
Adding or removing a substance to a system at equilibrium is a common application of Le Chateliers principle. Especially in industrial processes.
Rate Theory and concentration changes: Increasing the concentration of a reactant increases collisions which result initially in an increase in the rate of forward reaction but eventually as equilibrium is reached the forward and reverse reaction rates become equal. This is also true when the concentration of the product is increased but rate of reverse increases at first.
Note: The concentration in the solid/liquid phase does NOT change the rate because the surface area exposed is always the same therefore concentration remains constant.
Le Chatelier’s Principle and Temperature Changes
- Endothermic reaction: increasing the heat will shift to right for more products while decreasing heat results in shift left to make heat
- Exothermic reaction: increasing heat shifts to left to make more reactants while decreasing heat results in shift right to make heat
* Adding or removing energy from the system by heating or cooling will shift the equilibrium to minimize the change.
Le Chatelier’s Principle and Pressure and Volume Changes:
- Recall: Boyles Law = pressure is inversely proportional to volume. increasing pressure decreases volume and vice versa
- VOLUME CHANGES: by decreasing the volume of a gas will cause an increase in concentration. Thus changing the volume of a GAS may cause a shift in equilibrium. volume decrease results in shift to side with least amount of moles while increase in volume results in shift to side with more moles.
- PRESSURE CHANGES: by changing the pressure a shift in equilibrium may occur, however it is important to look at the TOTAL moles of reactants and compare them to the TOTAL moles of products of gas molecules. (note: pressure changes will invariably result in volume changes when dealing with gases).
Equilibrium will shift in a way to resists change, decreasing the volume of reactants shift the equilibrium to the right in order to reduce the pressure.
- sometimes no change occurs due to same amount of moles on either side –> Nothing will happen, equilibrium will not shift, since shift in neither direction will relieve pressure.
Note: Reactions with solids or liquids are not affected by changes in pressure since they are hard to compress.
the following stresses will affect equilibrium system:
Stress 1: Changing concentration
Stress 2: Changing temperatures
Stress 3: Changing volumes
Stress 4: Changing pressure
Stress 5: inert gases (cows) –> doesn’t change it just slows the reaction
Factors that do NOT affect position of equilibrium position:
1. Catalysts: Only decrease the time it takes for the reaction to reach equilibrium but does not affect the position or ratio. The role of a catalyst is to decrease the activation energy barrier for both the forward and reverse reaction.
- Adding inert gases: Adding a gas that does not take part in the reaction will not affect the equilibrium position. The most that happens is that a change in pressure is possible while volume remains constant. The role of an inert gas is basically decreases probability of collisions both reactants and products.
Solubility Eq’m
Terminology you should already be familiar with:
Unsaturated
Saturated
Supersaturated
Heterogeneous Equilibrium: A Solubility System
- When a sparingly soluble salt is added to water, ions at the surface of the solid enter the solution. Initially the concentration of ions is very low, therefore, the forward change is favoured and dissolving occurs:
MX(s) –> M+(aq) + X-(aq)
- As more ions enter the solution, the reverse change becomes more favourable and recrystallization occurs:
M+(aq) + X-(aq) –> MX(s)
- Eventually, the rate of dissolving equals the rate of recrystallization, therefore the system is at equilibrium:
MX(s) <–> M+(aq) + X-(aq)
@ eq’m rdissolution = precipitation
The Solubility Product Constant:
A saturated solution is one in which the solution is in contact with undissolved solute. As with any equilibrium, the extent to which dissolution occurs can be expressed by the magnitude of its equilibrium constant.
AbBa(s) <–> aA+(aq) + bB-(aq)
Ksp = [a]^a[b]^b
solids and liquids don’t appear in the eq’m expression
This equilibrium describes dissolving, therefore the equilibrium constant indicates how soluble the solid is in water and is referred to as the solubility product constant (Ksp).
Ksp equals the product of the concentrations of the ions involved in the equilibrium each raised to the power of its coefficient from the balanced equilibrium equation.
Solubility: is the maximum amount of solute that can dissolve in a given quantity of solvent at a particular temperature (saturated solution).
Example: silver chloride is slightly soluble ionic compound, a saturated solution will have three entities present at equilibrium: solid silver chloride, aqueous sliver ions and aqueous chloride ions.
Strong Electrolytes: are completely soluble in water (all alkali metals, nitrates, acetates)
Weak electrolytes: are slightly soluble in water (slightly dissolve and dissociate in water)
- if given the solubility of a solid this means that that much solid disolves into aq ions. create the balance equation including states of the disolving. write the equilibrium law Ksp based on coef. use the solubility and make sure in [] in M mol/L. understand the changes in the ice table and the solubility is a [] at equilibrium so use mole ratios to find it out or understand that the [] at 1 mol per litre is x. from there you can get the other eq’m values and can substitute into the equilibrium law to find Ksp. make sure the solubility is for only one mol of a substance and then use that value as x.
Recall: Solubility -the solubility of a substance is the maximum quantity of a solute that can dissolve in a given solvent.
- if given Ksp then fill in ice tables with changes of the aq ions. then sub into the Ksp equation and rearrange to isolate for x. then use x to find the solubility or eq’m [] which are same.
Precipitate reactions: two ways of establishing equilibrium between excess solute and an aqueous solution.
1. Dissolving excess salt into a solution until a saturated solution is reached.
2. Mixing two aqueous salt solutions together to precipitate the product out.
Will a precipitate form?
Recall: Q is the reaction quotient used to determine if a particular reaction is at equilibrium or not.
When considering solubility equilibria, the Qsp (the ion product) of a reaction can be calculated using any values that may or may not be at equilibrium. The significance of calculating the Qsp is that it enables one to depict whether a precipitate will form of the given ion concentration.
Compare Qsp to Ksp. if Q small than k then no ppt and unsaturated, if equal to k then no ppt but saturated and if Q greater than K then ppt and solution is supersaturated.
Steps to solving problems:
- Write the net ionic equation using solubility rules.
- find moles of each ion before mixing so c*v of the compound then use ratios to see if same moles for the ion.
- determine the total volume.
- Use new values (the moles of each ion and the total volume) to determine new concentration.
- Find the Qsp
- Compare Qsp to Ksp
- If Qsp > Ksp, therefore a ppt will form
The common ion affect
Adding an ion will shift the equilibrium, if the ion/compound reacts with the ions already present in solution.
ex. if you added NaF to MgF2 then it will cause a shift since the fluorine ions are same
it causes the solubility to decrease since there are more fluorine ions present now so it will shift to form reactant rather than ions so solubility decreases
when given equation set up ice table and make sure the concentration applies to the ion (like 0.01M of Ca(NO3)2 is 0.01M of Ca since its a 1 to 1 ratio when it ionizes), then write that [] in the initial of the ion with the other ion as zero and fill out rest of chart to solve for the x value with given Ksp. the x value is the molar solubility now.
Acid and Base Equillibirum
Acids: smaller ph than 7, sour tasting, indicators and red litmus and clear phenolphatlein, not slippery to touch and metals react by releasing H2 gas
bases: greater pH than 7, bitter tasting, indicator is blue litmus and pink pheolphathalein, slipper and no reaction with metals
H+1 is interchangeable with H3O+1 (called proton)
In an Organic acid, the hydrogen is at the end of the formula. In an Inorganic acid, the hydrogen is in front of the formula.
Svante Arrhenius definition of an acid and a base: Acid releases H+1 in water. Water takes on the hydrogen from the acid to form a hydronium ion (H3O+). In the textbooks, they sometimes refer to the hydronium ion as a proton, keep in mind that it is always a hydronium ion when dissolved in water. Base releases OH-1 in water
Bronsted-Lowry definition of an acid and a base:
- Acid an acid is a proton donor (and it does not have to be in water)
- Base is a proton acceptor.
- An acid always has a conjugate base and a base always has a conjugate acid. They always occur in pairs. A conjugate acid-base pair only differs by a proton.
- For reaction to occur: There must be an acid and base together. The products formed from this reaction differ by one proton and can in turn act as an acid or base. This is referred to as the conjugate acid or conjugate base depending on the products.
- According to the Bronsted-Lowry concept, acid-base reactions are universally reversible and result in an acid-base equilibrium.
(acid will lose proton or H becoming neg charges and called conjugate base while the base will gain proton or H becoming pos charged conjugate acid)
Substances like H2O that can act as an acid or a base are called AMPHIPROTIC (AMPHOTERIC).
A substance that can transfer more than one proton is POLYPROTIC
A strong acid has a weak attraction for protons while a strong base has a very strong attraction for protons.
The stronger the acid the weaker its conjugate base and the weaker the acid the stronger its conjugate base.
Strong Acids: Strong acid is an acid that ionizes quantitatively (completely) in water to form hydrogen ions. The percent ionization of a strong acid is approximately 99% or greater.
Strong Bases: All oxides and hydroxides of alkali metals. Alkaline earth metal oxides and hydroxides below beryllium (magnesium).
Water Equilibrium
Experiments have shown that pure, distilled water conducts slightly. This indicates that a small number of ions are present.
Autoionization of water:
H2O + H2O –> H3O+ + OH-
At SATP it was found that in one billion water molecules approximately TWO water molecules ionize. The reason why so few ions are present has to do with the molecules having the right orientation and energy.
It has been calculated that the concentration of water molecules in pure water or in a dilute aqueous solution is essentially constant. the water concentration value combines with the equilibrium constant of water to produce the ion product for water, Kw. Therefore Kw = [H3O+][OH-]
Water Equilibrium: based on the fact that very sensitive conductivity meters measured low readings for pure water, it indicates that a small number of ions are present.
Equilibrium equation for autoionization of water shows that there is a 1:1 ratio of hydroxide ions to hydrogen ions in pure water. At 25oC the concentration of both hydroxide and hydrogen ions is the same. From this the equilibrium constant for the ionization of water can be determined. found that the concentration of each is 1.010^-7 and the Kw is 1.010^-14. Note: if the temperature increases, then the Kw value increases since the equilibrium shifts to the right.
pH + pOH = 14 @ SATP
pH = -log[H3O]
pOH = -log[OH-]
[H3O+] = 10^-ph
[OH-] = 10^-pOH
Kw(1*10^-14) = [H][OH]
Acids ionize in the presence of water (covalent interactions)
Bases dissociate in the presence of water (ionic interactions)
Recall: Weak acids and weak bases slightly break down and set up equilibrium systems.
HX is strong acid while HA is weak acid
General term for expressing acid ionization, where HA represents an acid:
HA(aq) + H2O(l) –> H3O+1(aq) + A-(aq)
Equilibrium Law for acids:
Ka = [H3O][A-]/[HA]
BASES:
General equation for expression base reaction, where A-1 represents the base.
A-1(aq) + H2O(l) –> HA + OH-1
Equilibrium Law for weak bases:
Kb = [OH][HA]/[A-]
(if hydronium is on right side then acid rxn but if hydroxide on right side then base reaction)
Auto ionization takes place in all the aqueous solutions but when acids or bases dissolved in water then the hydronium and hydroxide ion concentrations are no longer equal.
Percent Ionization of weak acids:
The more dilute the solution of a weak acid, the greater the degree of ionization (percent ionization). Think about it, good old Le Chatelier’s principle says adding more water to an equilibrium system would cause a shift to the right making more of the ionized acid species. The addition of water would change the concentration of the ionized ions thus Le Chatelier will seek to reestablish those ion concentrations by shifting to the right.
Percent ionization (p) = concentration of acid ionized x 100% Concentration of acid solute
or %p = [H+]/[HA] *100%
ex. if given pH and molarity set up ice or if then find the eq’m values or use pH to find [] then plug in the [] into %p to find
Ionization of strong acids is almost 100%. Therefore, the concentration of the hydronium ion and its anion will be stoichiometric. (if)
The difference between a strong acid and a weak acid is that a weak acid does not ionize completely in water (<50%) to form H+1. Most common acids are weak (aspirin, citric acid, vinegar, vitamin C). Due to the incomplete ionization equilibrium will be set up. Thus, the methods learned earlier solving equilibrium problems will be used exactly in the same way, except this time it involves weak acids.
Steps to follow when considering weak acids:
Is this a weak acid or strong acid? Look at the Ka value. Strong acid will have a large value and a weak acid will have a small value. (then set up ice or if)
Write the major species in solution for the weak acid. (The weak acid will predominate as compared to the ionized species)
Compare the species of water to the weak acid by looking at the Ka versus Kw values. This is important because it is possible for the acid to be so weak that the contribution of hydrogen ions from water auto-ionization to be greater. If the hydrogen ion concentration of water is much less than the weak acid then it can be ignored. If the hydrogen ion concentration of water is greater, then it must be taken into account during calculations. (if greater than 1*10^-14 then set up ice but if smaller then pH of 7)
Make sure your chemical equation is balanced.
Set up an ICE table.
Solve for x. (see if you can make your life easy by making a valid assumption)
Determine the concentration of unknown species.
Solve for any pH using [] found
see if q=Ka
Strong Bases and Weak Bases
Strong bases like KOH, NaOH, LiOH, Ca(OH)2, Ba(OH)2, Sr(OH)2 undergo complete dissociation in water. these are if questions/stoich questions
Note: A base does NOT have to contain a hydroxide ion. For example, ammonia is a base because it produces hydroxide ions when dissolved in water. Any species containing a nitrogen atom with a lone pair of electrons will exhibit base-like properties.
Weak bases do not undergo complete dissociation in water and thus set up equilibrium system.
Note: Kb refers to the reaction of a base with water to form the conjugate acid and the hydroxide ion. Strong bases will have very large Kb (therefore quantitative) values while weak bases will have small Kb values (same process as weak acids, make sure greater than Kw, set up ice table, solve for x in equation, use [] values to find pH etc.)
ACID-BASE PROPERTIES OF SALTS
Usually, salts (ionic compounds) breakdown in water to make independent ions that do not really react. Under certain conditions some ions have the ability to behave as acids or bases.
Salts That Produce Neutral Solutions:
Important rule to remember: salts that consists of the cations of strong bases and the anions of strong acids have no effect on the [H+] when dissolved in water. And so, aqueous solutions of salts contain anions and cations of strong acids and bases are neutral, pH=7. Examples: Cl-, NO3-1, K+1, Na+1. if both cation and anion come from strong acid and base the ions are both weak conjugates and don’t ionize in water so pH still 7
Salts That Produce Basic Solutions:
Rule: For any salt whose cation has neutral properties (Na+1, K+1) and whose anions is the conjugate base of a weak acid, the aqueous solution will be basic. (strong conj base will ionize water making it basic)
Ka x Kb = Kw
ex. given the molarity of a solution and Ka or Kb figure out if coming from strong or weak acid or base to see if acid or basic salt. then use ice table to get the eqm [] and see if its kb or ka and from there see if ka or kb given if opp given then do kw/kgiven to find ur ka or kb value to solve for x
Salts That Form Acidic Solutions
Rule: A salt whose cation is a conjugate acid of a weak base produces an acidic solution. A second type of salt that produces an acidic solution is one that contains a highly charged metal (Al+3, Fe+3,Sn+2) ion. Typically, the higher the charge on the metal ion, the stronger the acidity of the hydrated ions.
for stuff like al or metals with high charge they are likely a strong conj acid so they create acidic solutions. its usually [Al(H2O)6] + H2O –> [Al(H2O)5OH] + H3O and its ka.
Acid-Base Titrations
Purpose of titration: determine the unknown concentration of a solution using a known concentration of a different solution.
Titration: chemical analysis involving the stepwise addition of a solution with a known concentration (titrant) into a measured volume of a solution with an unknown concentration (sample). A titration experiment is carried out several times to ensure some consistency, however, it is virtually impossible to achieve a high degree of precision due to experimental errors.
Materials needed for an acid-base titration:
Burette (the titrant)
Pipette with bulb
Flask (sample or unknown)
Appropriate indicator
Titrant
Sample
Crucial terminology:
Acid-base titration: the determination of the unknown acid or base concentration involving a neutralization reaction.
Equivalence point: chemically equivalent points have reacted (number of moles of H+1 = number of moles of OH-1). (nH = nOH)
Endpoint: colour change of an indicator during titration.
Acid-Base Indicator: signals the end of titration by sharply and permanently changing colour when the equivalence point is reached. Ideally, the selection of the acid-base indicator is done such that the end point occurs precisely at the equivalence point of the reaction.
What are acid-base indicators: weak monoprotic organic acids. Different types of indicators change colour at different pH’s. The colour change occurs over a 2 point pH range. Refer to Fig 9.3 on p.425 for a list of indicators and their pH.
(strong acid + strong base will give if only)
(weak acid and strong base will give if for the [] but ice for pH and the conj salt ionizing water)
Titrating a Strong Acid with a Strong Base: the pH will be 7. there are 4 points –> 1. is before any base (if to get pH). 2. 1/2 way point to neutralization –> you have to use if again to get the [] and then find pH. 3. is equivalence point there’s no acid or base just the salt and pH will be 7. 4. is above and beyond equivalence point and u use stoich to see how the excess will affect the pH (pH = 14-pOH)
Titration of a Weak Acid with a Strong Base: this time the solution will be basic since the strong conj base of the salt is present at the equivalence point. 1. you find the pH of the acid using ice since its a weak acid. 2. at halfway point you do if with the concentrations and volumes of each to get initial moles and then concentrations from the total volume. then use ice of the acid breaking in water to find the pH. 3. once again use if then ice except with the new volumes to find the pH. 4. find the beyond pH by using if of the extra concentration of the strong base which makes the solution basic.
In general, the pH at the equivalence point is higher than 7 for a weak acid with a strong base.
Titrating a Weak Base with a Strong Acid: pH will be smaller than 7 since the salt will have a strong conj acid. Once again you do the same calculations. 1. use ice to find the pH of the weak base. 2. use if to find the [] of entities leftover then ICE to get the pH. 3. use if and ice with the new values. 4. if likely if to get the pH.
In general, the pH at the equivalence point is lower than 7 for a weak base with a strong acid.
for strong acid and strong base use if throughout
for a weak something and strong something these are the steps of calculating pH for a pH curve:
1. first use ICE to get pH of the solution that is going to be titrated by a strong something. use ICE of the weak substance ionizing in water and find out what x is then use x to get the pH.
2. now use IF with the reaction of the weak and strong thing reacting. use the initial concentration, volume and multiply to get moles. the strong substances is limiting until equivalence point and all moles are used. in final the moles of the weak susbatcne is initial - the initial moles of strong substance. then use to total volume to get the final concentrations of the weak substance and the salt. now use ICE of the weak substance ionizing in water by using the final concentration values used (including common ion) and finding what x is and getting pH.
3. now use IF again of the weak and strong reaction. the initial moles of both should be equal now or at least the volume of the strong substance should be all of it. then get final knowing that the final moles of the reactnats is zero and only final moles of the salt exists. then get the concentration of it. Now use ICE of the strong conj whatever of the salt ionizing in water. initial concentration is one found in IF and use it to get x and then find pH
4. if going overboard on the titrant/strong substance use IF of the acid and base reaction to get the moles. the final moles of weak is zero while for the titrant and salt it exists and now find the concentrations of them. since the strong titrant is in the solution it will affect the pH and now pH is calculated using its moles (if one to one ratio) to get the pH.
Polyprotic Titrations
Recall: polyprotic acids can donate more than one hydrogen, but bases can also be polyprotic. For example, carbonate is a base that can accept two hydrogen ions.
But remember polyprotic acids Ka values show that the dissociation of a hydrogen becomes harder with every proton being donated. And this is also true for a base.
Conclusion: since more than one hydrogen can be donated (acid) or accepted (base), in a titration reaction more than one end-point is detected.
General rule: only quantitative reactions produce detectable end-points in an acid-base titration. (the step in the dissociation with the biggest Ka or Kb value).
Buffers
pH curves involving a weak acid or a weak base have at least one region where buffering occurs. A buffering region on the pH curve is where there is very little change in pH despite the addition of an appreciable amount of (strong) acid or (strong) base. It usually occurs at half the volume of the equivalence point.
Two ways of preparing a buffer to produce an equal molar concentration of a conjugate acid-base pair:
Mix a weak acid with the salt of a conjugate base.
CH3COOH + NaCH3COO
Mix a weak base with a soluble salt of a conjugate acid.
NH3 and NH4Cl
(basically what u need is a weak acid and conj salt (that is stong base) or weak base and conj salt that is strong acid)
Another way of preparing a buffer is by conducting a neutralization reaction of a weak base or weak acid, but the weak base or weak acid must be in excess, such that buffer conditions are present (conjugate acid-base pairs)
Example are below:
Mix 0.1M, 100mL ammonia with 0.1M, 50.0mL hydrochloric acid. (use if to get final concentrations, if same then buffer is generated and there is equal [] of the acid and salrt or base and salt)
How does a buffer work?
Example: acetic acid and sodium acetate
Addition of an acid:
adding H+ would result in a shift left to make more of the acid rather than the ions
Addition of a base:
addition of OH results in the H+ reacting with it and a shift to the right to make more ions rather than the acid
Will a buffer keep working forever? No.
There is a limit to how much acid or base that can be added to a buffer (for neutralization) before its pH begins to rise rapidly. This limit is known as Buffer Capacity. The extent of the buffer capacity is determined by the concentrations of its conjugate acid-base pair. The higher the concentration of a buffer, the higher its resistance of pH changes and vise versa.
Buffer capacity is defined as the amount of strong acid or base needed to change the pH of a 1.0 L of a buffer solution by ONE unit.
in a problem where it says blank amount of buffer is prepared that contains those much acid or base and this much conj salt at equillbiurm and asks for the pH and then the pH when amount of acid or base added what you do is you first use ice of the acid or base dissolving in water and use the [] of the acid or base and the common ion in salt, then find the x and find pH. now you you find the actual moles of the acid or base and the salt and understand what you are adding. create a reaction of the acid or base ionizing and see what would increase and decrease and use the moles and add or subtract that amount. now us e the moles to get back their concentrations and input them into a ice table of the acid or base ionizing and solve for x and then pH, it should show a minimal change in pH indicating that the buffer is working.
