Acid Base Equilibria. Flashcards
What is the Bronsted lowry definition of an acid and also a base.
An acid is a substance that can donate a proton. A Base is a substance that can accept a proton.
What is an alkali defined as.
A water soluble base that produces OH- ions in aqeous solution.
What is different about strong acids and weak acids.
Strong acids will fully dissociate in solution to form the maximum number of H+ ions. Weak acids will only partially dissociate into H+ ions.
What is special about the {H+} of a strong monoprotic acid.
The concentration of [H+] is the same as the concenration of the acid, [HA}. Disociate in a 1:1 ratio.
How must acids and bases react.
They must recact in pairs, one acid and one base. Each acid is linked to a conjugate base on the other side of the equation always.
How can you easily identify conjugate acid and base pairs.
Need to think about what has potential to donate a H+ ion. Potential is the key concept. Potential is in terms of could it take a H+ from its conjugate pair to form the form the substance that the other compound was originaly.
Give the conjugate acid and base pairs in
1) Hydrogen chloride plus ammonia going into ammonium chloride.
HCL is acid Cl- is base ,consider the ions sepertely if there is only one ionic product to give two needed products.
NH3 is a base and NH4+ is an acid.
Give the conjugate acid and base pairs in HCl + H2O going into H3O+ + CL-
And name the ion that is produced.
HCl is a an acid and Cl- is the conjugate base. Pair
H20 is the base and H30+ is the acid. Pair
It is known as the hydroxonium ion.
What acuracy should you give pH to.
Always 2 decimal places.
Define pH.
Give this exact definition nothing else
pH= -log10 [H+(aq)]
How do you find the concentration of H+ ions in Moldm-3 given the pH. How is the concentration of H+ in solution represented.
Take the antilog. Just do 10 to the power of the negative pH. The square brackets just means concentration of the entity in moldm-3.
What equilibrium is established in all aqeous solutions an pure water.
H20(l) reversable reaction into H+(aq) and OH- (aq)
Why does the H+ ion have strange chemical properties compared to other possitive ions.
It has no electrons just the hydrogen nucleus. Its diameter is much less than any other chemical entity, this extreamly small diameter means that it has an intense electric field.
In reality how is H+ found in solution and how is H+ actualy represented for the sake of simplicity.
H+ in reality is actually always bonded to at least one water molocule and so is found as H3O+ however it is just represented as H+ for simplicities sake.
Since H+ has no electrons of its own what is the only type of species that the H+ ion is able to bond with.
It is only able to dative covalent bond with species that have a lone pair of electrons.
What Kc Expression can be created from the equilibrium established in water and aqeous solutions.
Kc = [H+(aq)} {OH-(aq)}
{H20(l)}
Considering the equilibrium that is established in water how can this be rearanged to give soemthing usefull and state the asumptions you made to get there.
The equilibrium established in water and aqeous solutions can be rearanged to give Kc x [H2O} = {H+} {OH-}
You asume that since {H2O] is much larger than the ions you asume that since so little of it dissociates it remains constant. H20 has stong covalent bonds so little disocaites.
You can hence ignore this from th expression but since you have done so it can be called Kc beacsue you just got rid of a term.
It is instead called the ionic product of water- Kw.
Write an expression for the ionic product of water Kw and state why Kw is practically usefull in calculation.
Kw= {H+(aq)} {OH-(aq)] Must know off by heart.
If you need to convert between {H} and {OH} you just take Kw as 10 to the power of negative 14 (unless another value given for another temp) and re-arange. Divide Kw by either.
Note that the values must be concentrations in moldm-3 to go into the expression.
What is the value for Kw including units, at 298K, 25 celcius.
10 to the -14 Mol2dm-6
Why is pure water considered a neutural solution.
It is considered a neutural solution since a neutural solution is defined as one where {H+(aq)} = {OH-(aq)]
Define a neutural solution.
One where {H+(aq)} = {OH-(aq)]
Since Kw is a constant at a given temperature what must happen to {OH-}eqm if [H+] eqm is increased and vice versa.
If the concentration of one of thease ions increases the the concentration of the other must decrease proportionally in order to keep Kw constant.
What can be said for the Kw expresion in all neutural soloutions such as water.
Kw = [H+} squared (very different from modified ka)
Write a proof that the pH of pure water will be 7 at 25 degrees celcius.
SEE PIC square root ka and sub into pH expression.
How will the pH of pure water change with increased temperature.
The disociation of water is an endothermic process. In an exam you would write the equation for this. Since energy is required to break luiqid water bonds to form aqeous ions the forward reaction is endothermic. Increasing the temperature would shift the pH to the RHS favoring a forward reaction to resist the temperature increase, and hence the degree of dissociation will increase and {H+} will increase.
So pH will decrease.
How is pH and neutrality related. Can water be neutural if the pH doesnt =7.
Need to understand that pH is a measure of [H+} and is -log10[H+(aq)] numerical value. A neutural solutuion is one where {H+} = {OH-} and so NEUTURAL SOLUTIONS DONT ALWAYS HAVE pH=7. It is possible for water to still be neutural with a pH that isnt 7 if by definition {H+} = {OH-}.
Why is a logarithmic scale used for pH even though it is more complex than just stating the number of H+ ions.
The concentration of H+ ions in most solutions is actualy very small. So would give very small values out. The pH scale will give almost all possitive numbers since the log of anything less than one will be negative and so by X that by -1 you get possitive non awkward values, doing away with tiny values like 10 to the negative 3. normally between 0-14 , but can be outside this. Decimals are possible.
State cases where the pH of a solution would be less than 0 and also greater than 14.
Would be less than 0 , negtive , if the solution has a [H+] of greater than 1 moldm-3. Since you would esentialy be raising 10 to a power than would get you a possitive number so this would have to be a possitive fraction. When this is timesed by -1 it will turn negative.
A 1M H+ solution will have a Ph of zero.
A 1M OH- solution will have a PH 14
Going over 1 either will mathaematically breaks such boundries.
What does a decrease in 1 mean for the change in [H+].
It has increased by a factor of 10.
Give a detailed method for constructing a Titration Curve.
1.Transfer 25cm3 of acid to a conical flask with a volumetric pipette2 .Measure initial pH of the acid with a pH meter3 .Add alkali in small amounts(2cm3) noting the volume added 4.Stir mixture to equalise the pH5. Measureand record the pH to 1d.p.6. Repeat steps 3-5 but when approaching end point add in smaller volumes of alkali 7 .Add until alkali in excess.
State one way you could improve acuracy when experimentally constructing a pH curve.
By keeping a constant temperature.
State how you would calibrate a pH meter and why it is neccasary to do so.
Calibrate meter first by measuring known pH of a buffer solution. This is necessary because pH meters can lose accuracy on storage. Most pH probes are calibrated by putting probe in a set buffer (often pH 4) and pressing a calibration button/setting for that pH.Sometimes this is repeated with a second buffer at a different pH.
If you where given a tritration curve or needed to draw one how would you find the neutralisation volume and wht data will be relevent for this.
You can find the neutralisation volume by way of simple titration calculation. You need to find the volume of whatever is added from the beurete. If they give you a value then just consider this as the amount you have available, carry out a titration calculation as normal and find your own volume by considering it as unknown as you are concerned with the amount required to neuurtalise (recact with all the moles) of what is in the conical flask.
What is known about strong bases. Show this via an equation.
Strong bases completely dissociate into their ions. NaOH into Na+ + OH-
Regarding most strong base calculations how would you find the pH of an isolated strong base.
For bases we are normally given the concentration of the hydroxide ion. To work out the pH we need to work out [H+(aq)] using the Kw expression taking the Kw to= 10 to the minus 14 if at 25 celcius.
Calculatethe pH of the strong base 0.1mol dm-3 NaOH , assume complete dissociation.
Answer is 13.00. Solution on example 3 in notes.
What would you need to do for a strong base calculation that involved 2OH ions in the molocule such as Ba(OH)2.
Take the concentration of the base in M and multiply by 2 to find the concentration of OH- since it will dissociate in 1:2 ratio , each molocule giving way to 2 Ions when in aqeous solution.
Calculate the pH of 20grams per dm3 of NaOH
13.7
Calculate the pH of solution formed when 50cm cubed of 0.25M KOH is made up to the mark of 250cm3 by adding water.
12.7
When diluting an acid or an alkali what formula should you use and what from this formula is needed to go into any expression.
pH of diluted strong acid method
[ H+]new= [H+] old x old volume
———————
new volume
Same method if using OH- but you will need to put the [OH-] new into the kw expression to get conversion for {H+}.
You can only put the new diluted concentrations into expressions to calculate pH value.
Calculate the new pH when 50.0 cm3 of 0.150 mol dm-3. HCl is mixed with 500 cm3 of water
1.87 .taken from example 13 on revision notes.
Calculate the pH of 0.2M of Ba(OH)2 .
13.6
What must you know about weak acids. As always use an equation to support your answer.
Weak acids only slightly dissociate when dissolved in water, giving an equilibrium mixture.
HA + H2O (l) REVERSABLE INTO H3O+(aq)+ A-(aq)
Can be simplified into more usefull.
HA(aq) REVERSABLE INTO H+(aq)+ A-(aq)
Write an expression for the weak acid dissociation constant, Ka. Do this in both general terms and also for propanoic acid giving the equation for the dissociation of propanoic acid.
General : Ka= [H+(aq)][A-(aq)]
[HA (aq)]
For weak propanoic acid:
CH3CH2CO2H(aq) Reversable into H+(aq) CH3CH2CO2-(aq)
So........
Ka= [H+(aq)][CH3CH2CO2-(aq)]
[CH3CH2CO2H(aq)]What does a given value for Ka tell you about the strength of an acid and why.
The larger the Ka the stronger the acid. A stronger acid will dissociate to a greater extent this means that the magnitude of [H+(aq)][A-(aq)] will be greater and hence the top half of the ka “fraction” will be greater.
State what scenario would require you to modify your Ka expresion for the purpose of making calculation possible.
When given an isolated weak acid with too many unknowns. You will be given either the concentration of the isolated weak acid or its pH and asked to calculate the one you have not been given.
What two asumptions are made to modify the Ka expression and state what the new modified Ka expression actually is.
1)[H+(aq)]eqm= [A-(aq)]eqm because they have dissociated according to a 1:1 ratio.
2) As the amount of dissociation is small we assume that the initial concentration of the undissociatedacid has remained CONSTANT . So[HA(aq)]eqm= [HA(aq)]initial
Since its weak and will dissociate a small amount we assume that none has dissociated at all.
Ka = [H+(aq)]2
[HA(aq)]initial Where Ka is given and [H+(aq)] can be found from the antilog of pH.
Example 5: What is the pH of a solution of 0.01 mol dm-3ethanoic acid (ka is 1.7 x 10-5mol dm-3)?
pH = 3.38 (about right for a weak acid)
Use the simplified ka expression.
