Acids, Bases and Buffers Flashcards
What are the the bronsted-lowry definition?
An acid is a proton donor.
A base is a proton acceptor
Why is HCL a strong monobasic acid?
• Strong Acid – It fully dissociates into H+ and Cl− when dissolved in water.
• Monobasic – one mole of acid releases one mole of H+ ion
What is an example of a dibasic and tribasic acid?
-H2SO4
-H3PO4
What is an equation for the combination of H2O with an H+ ion?
H+ + H2O—> H3O+
what type of bond forms when h+ and h2o combine?
dative covalent bond
What are HCl and Cl called in this equation- HCl(aq) → H+ (aq) + Cl- (aq)?
HCl and Cl- are called a conjugate acid—base pair
What is a conjugate acid-base pair
two species that can be interconverted by transfer of a proton
In this equation- HCl(aq) → H+ (aq) + Cl- (aq), what happens?
• Here, HCl donates the proton and Cl- is the conjugate base.
• In the reverse reaction Cl- can accept a proton to form the conjugate acid, HCl.
How would I explain this equation? CH3COOH (aq) + NH3(aq) CH3COO- (aq) + NH4+ (aq)
Acid 1 Base 2 Base 1 Acid 2
CH3COOH is an acid (1) as it donates H+; in the reverse reaction CH3COO- is the
conjugate base (1) as it accepts H+
• NH3 is a base (2) as it accepts H+; in the reverse reaction NH4+ is the conjugate
acid as it donates H+
What does the reaction between magnesium carbonate with dilute nitric acid form?
MgCO3(s) + 2HNO3(aq) → Mg(NO3)2(aq) + CO2(g) +H2O(l)
make sure to practice ionic equations
What is the difference between strong and weak acids?
A strong acid completely dissociates into its ions in water
HCl(aq) → H+ (aq) + Cl- (aq)
Whereas a weak acid is only partially dissociated into its ions in water
CH3COOH(aq) CH3COO- (aq) + H+ (aq)
Organic acids are weak acids, not all inorganic acids are strong.
Ethanoic acid reacts with an excess of magnesium carbonate. Write a
balanced equation for the reaction including state symbols.
2CH3COOH(aq) + MgCO3(s) (CH3COO)2Mg(aq) + CO2(g) + H2O (l)
What ion is common to all acidic solutions?
H+
50.0 cm3 of 1.00 moldm-3 HCl reacts with excess powdered MgCO3. The volume
of CO2 produced is 0.600 dm3
50.0 cm3 of 1.00 moldm-3 CH3COOH reacts with excess powdered MgCO3.
The CO2 gas is given off at a lower rate.
Use your knowledge of theory of reaction rates and dynamic equilibrium to explain why CH3COOH(aq) reacts with MgCO3 at a slower rate than HCl
(aq)
HCl H+ + Cl−
• HCl is a strong acid and is fully dissociated.
• Eqm lies very far to the RHS and [H+] is high.
• CH3COOH H+ + CH3COO−
• CH3COOH is a weak acid and is partially dissociated.
• Eqm lies to the LHS and [H+] is low.
• [H+] is lower, less frequent collisions, less frequent successful
collisions per unit time.
50.0 cm3 of 1.00 moldm-3 HCl reacts with excess powdered MgCO3. The volume of CO2 produced is 0.600 dm3
50.0 cm3 of 1.00 moldm-3 CH3COOH reacts with excess powdered MgCO3.
The CO2 gas is given off at a lower rate.
State and explain if the volume of CO2 produced in the second experiment
would be more, less or the same?
• Volume of CO2 will be the same.
• H+ ions react with MgCO3, eqm shifts to the RHS to increase [H+].
• More CH3COOH will dissociate until all of the acid has reacted.
• Moles of each acid are the same (volumes and concentrations
are the same)
What is Ka?
Is a measure of the extent of dissociation of an acid.
A stronger acid will have larger Ka values but weak acids have relatively much
smaller.
What is the equation for Ka (weak acid)?
= [H+ ] [A- ] /[HA]
What is the equation of pKa?
pKa = - logKa
The inverse operation is also useful Ka= 10-pKa
What does a lower Ka mean?
weaker the acid
What does a lower pKa mean?
stronger the acid
What is the pH definition?
pH = −log [H+]
used to find ph of strong acid
For a strong monobasic acid (fully dissociated) what is will the [H+(aq)] be the same as?
the concentration of the acid.
What are pH values normally quoted to?
2 dp
What is pH best determined by?
using a pH meter rather than indicator paper
How does we find the pH of a string acid when dilution has occurred?
We can use a dilution factor or amounts of substance to calculate a new
concentration and then work out the new p
A student added an equal volume of water to 25.0 cm3 of 1.00 mol dm-3 HCl.
Calculate the pH of the new solution.
• Dilution factor = 2 (the total volume would now be 50cm3)
• The dilute acid has a concentration of 1.00/2= 0.500 (it is 2 times less
concentrated)
• pH = -log 0.500 = 0.30
How do you work out the pH upon reacting a strong acid with a strong base?
Full equation
Ionic equation
n(h+)
n(base)
n(h+) remaining
new vol
H+ (n/vol)
ph
What is HA?
weak acid
for a weak acid not fully dissociated what can’t we say?
the eqm conc of protons is the same as that of the acid
What is [H+] equal to?
[A-]
What is the re written expression for Ka?
Ka = [H+]2/
[HA]
How is [H+] found?
√ (Ka x [HA])
Write the Ka expression for butanoic acid (CH3CH2CH2COOH). Find the pH of a 1.00 mol dm-3 solution, Ka = 1.51 x 10-5 mol dm-3
1.00 mol dm-3 solution, Ka = 1.51 x 10-5 mol dm-3.
Ka = [CH3CH2CH2COO-][H+]
[CH3CH2CH2COOH]
If we assume that Ka = [H+]2 then
[HA]
[H+] = √ (Ka x [HA]) = √ (1.51 x 10-5 x 1.00) = 3.86 x 10-3 mol dm-3
pH = -log [H+] = -log(3.86 x 10-3) = 2.41
What is the acid dissociation constant?
pKa
How to work out % dissociation?
h+/ha x 100
HA is a weak carboxylic acid found in the body.
The Ka value of HA is 1.38x10-5 mol dm-3
A student analyses a sample of HA using the procedure outlined.
• Dissolves 0.0113g of HA in water and makes the solution up to 0.500 dm3
• The student measures the pH of the resulting solution as 4.23
Determine the molar mass of HA, molecular formula and suggest a possible structure
for HA
HA has one carboxylic group and one OH group for an alcohol and contains C, H and O only.
Show all your working
[H+] = 10-4.23 = 5.8884 x 10-5 mol dm-3
Ka = [H+]2/[HA]
HA= (5.8884 x 10-5)2/1.38 x 10-5 = 2.51258 x 10-4 mol dm-3
n = cv = 2.51258 x 10-4 x 0.5 = 1.25629 x 10-4 moles
n = m/ mr Mr = 0.0113/1.25629 x 10-4 =89.95 ≈ 90 g mol–1
COOH = 45 g mol–1
OH = 17 g mol–1
Total: CH2O3 = 62 g mol–1
Left: C2H4 = 28 g mol–1
Mol. Formula C3H6O3 = 90 g mol–1
what is the dissociation of water equation
H2O <=>H+ + OH- ΔH= +v
Explain why ΔH is positive?
It is an endothermic process as energy is being absorbed in order to break
bonds.
At room temp was is H+ and ph?
[H+] =1.00 x 10−7 mol dm-3, so the pH is 7.00.
At other temps why is H+ diff?
because the equilibrium will have
shifted, however, water can still be neutral even if the pH is not 7.00.
• It is neutral because [OH–] = [H+]
What is the dissociation equation for kw?
Kw = [H+][OH-]
At room temperature what is H+ and OH?
[H+] = [OH-] = 1.00 x 10-7 mol dm-3
What is the value and units of Kw?
1.00 x 10-14
mol2 dm-6
For PH values that are whole numbers l why can you work out the H+ OH- concentrations?
as the indices for H+ OH- add up to -14
When is water neutral?
when [H+] = [OH]
How do we find the pH of a strong base?
[H+] = Kw/[OH-]
Finding Ph upon reaction of a strong acid with a strong base- Worked
During a titration a student added 10.0 cm3 of 0.100 mol dm-3 HCl, from a burette to 25.0 cm3 of 0.100 mol dm-3 NaOH in a conical flask. Determine the pH of the solution formed if the experiment was carried out at 298K.
Full equation: NaOH + HCl—>NaCl + H2O
Ionic Equation: OH– + H+ —> H2O
n(HCl ) = 10.0 x 0.100 = 1.00 x 10-3 mol
1000
n(H+) = 1.00 x 10-3 mol
n(NaOH) = 25.0 x 0.100 = 2.50 x 10-3 mol
1000
n(OH-) = 2.50 x 10-3 mol
n(OH-) remaining = 2.50 x 10-3 - 1.00 x 10-3 = 1.50 x 10-3 (alkali in excess)
New volume = 10.0 + 25.0 = 35.0 cm3
[OH-] = 0.0015/ 35.0 x 1000 = 0.04286 mol dm-3
[H+] = Kw = 1.00 x 10-14/0.04286 = 2.33 x 10-13 mol dm-3
[OH-]
pH = -log10[H+] = -log10[2.33 x 10-13] =12.63
what is a buffer solution?
one that minimises changes in pH when small amounts of acid or alkali are added
In what systems is it important to keep the pH constant in?
• Shampoo safety
• Effectiveness of blood in transporting oxygen
• Stability of molecules in living systems
• Effectiveness of chromatography in separating mixtures
what do buffers contain?
-two species that “remove” the added small amounts of acid or alkali
added.
-Species 1 is a weak acid (HA) and species 2 is its conjugate base(A-)
What does a buffer solution do minimise the changes in pH?
Here we will consider a mixture which contains ethanoic acid and sodium
ethanoate.
The acid partially dissociates because it is weak.
CH3COOH <=> CH3COO– + H+
• The salt is a source of the conjugate base which fully dissociates so we get a lot of ethanoate ions.
• The result is that we have large amounts of both ethanoate ions (conjugate base) and of ethanoic acid in the mixture.
• ethanoate ions and of ethanoic acid can counteract the addition of acid or
alkali.
If we add acid, the H+ ions react with the ethanoate ions to make ethanoic acid
Equation 2:
CH3COO– + H+ <=> CH3COOH
Equilibrium in Equation 1 to shifts to the left, removing most of the H+ ions.
If we add OH- ions, they react with H+ ions
Equation 3:
OH– + H+ H2O
Equilibrium in Equation 1 to shift to the right.
(Or we could say they react with CH3COOH to form the conjugate base and water).
CH3COOH + OH– CH3COO– + H2O
net effect is that the pH only change very slightly because we have a large
reservoir of ethanoate ions and ethanoic acid to ‘absorb’ any acid or alkali added.
There is a limit to how much acid or alkali you can add.
What is the summary for the general action of a buffer?
-Write this equation for the acid concerned:
HA H+ + A- [1]
Addition of an acid
Equilibrium will shift to the left The added acid reacts with A-
Or you can write this as an equation H+ + A- → HA
Addition of an alkali
Equilibrium will shift to the right The added alkali reacts with H+ (or the named acid)
Or you can write an equation H+ + OH- → H2O
what are the methods used to make a buffer?
-mixing a weak acid with its salt (CH3COOH (acid)+ CH3COONa (salt)
-partial neutralisation of a weak acid- СН3СООН + NaOH (SB)→ CH3COONa (Salt) + H2O
WA in excess
n(salt) formed = n(SB)
n(WA) left = n(WA) - n(SB)
How is a buffer made in the second method?
buffer is made by adding excess weak acid to a solution of an
alkali like sodium hydroxide.
In this way you end up with the same mixture as in a) a solution containing the weak acid and the salt of the weak acid.
What is the Ka expression used to calculate the pH of a buffer?
[H+ ]=Ka x [HA]/[A-] (
In a buffer solution why does [H+] not equal [A-]?
because the [A- ] has been added separately as one of the components.
What does it mean If the buffer contains a 1:1 ratio of [HA]:[A−]?
the pH = pKa, because [H+] = Ka
Calculating pH of a Buffer Solution: Mixture of a Weak Acid and its Salt
H+] = Ka x [HA]/[A-] = 1.70 x 10-5 x 0.500/0.100= 8.50 x 10-5 mol dm-3
pH = -log[H+] = -log(8.50 x 10-5) = 4.07
What is anything ending in ate?
salt
What happens at half neutralisation?
n(WA)=n(salt)
Ka=[H+]
pKa=pH
what would happen to the pH of a solution Upon addition of potassium butanoate?
[H+] = Ka x [HA]/[A]
[A-] increases
[HA]/[A] decreases
[H+] decreases pH increases
What salt would needed to be added to phenol in order to make a buffer solution?
phenol with o- instead of oh
Calculating Reaction of excess WA with a strong base-What is the pH of a buffer solution if 25.0 cm3 of 0.500 moldm-3 sodium hydroxide is
mixed with 75.0 cm3 of 0.500 mol dm-3 propanoic acid at 298K. The Ka value for
propanoic acid at this temperature is 1.30 x 10-5 mol dm-3
-Initial n(CH3CH2COOH) = 75.0 X 0.500/1000 = 0.0375 mol
Initial n(NaOH) = 25.0 X 0.500/1000 = 0.0125 mol
CH3CH2COOH + NaOH → CH3CH2COONa + H2O
initial moles 0.0375 0.0125 0 0
change in moles -0.0125 -0.0125 +0.0125 +0.0125
final moles 0.0250 0 0.0125 0.0125
Volume terms within [HA] and [A−] cancel out mathematically, n(HA) and n(A−) can
be used to calculate [H+]
Hence calculate the pH of the buffer solution:
[H+] = Ka x [HA]/[A-] = 1.30 x 10-5 x 0.0250/0.0125
= 2.60 x 10-5 mol dm-3
pH = -log[H+] = -log(2.60 x 10-5) = 4.59
-Calculating pH of a Buffer Solution:
After Addition of a Base to a Buffer Solution
A buffer solution contains 0.200 mol of propanoic acid and 0.180 mol of sodium
propanoate in 1000 cm3 of solution. A 0.0150 mol sample of solid sodium hydroxide is then added to this buffer solution. Ka for propanoic acid = 1.30 x 10-5 mol dm-3
Calculate the pH after addition of the base.
Calculate the number of moles of propanoic acid and of propanoate ions present in the buffer solution after the addition of sodium hydroxide.
CH3CH2COOH + NaOH → CH3CH2COONa + H2O
initial moles 0.200 0.0150 0.180
change in moles -0.0150 -0.0150 +0.0150
final moles 0.185 0 0.195
Hence calculate the pH of the buffer solution after the addition of the sodium
hydroxide.
[H+] = Ka x [HA]/[A]
[A-] increases
[HA]/[A] decreases
[H+] decreases pH increases
What does moles left equal using a buffer adding strong base?
n(WA) left- initial n(WA)-n(SB)
n(salt) left- initial n(salt) + n(SB)
What does moles equal using a buffer adding strong acid?
n(WA) left- initial n(WA)+n(SB)
n(salt) left- initial n(salt)-n(SB)
- A student adds 50.0 cm3 of 0.500 mol dm-3 butanoic acid to 25.0 cm3 of
0.250 mol dm-3 sodium hydroxide. A buffer solution forms.
(a) Explain why a buffer solution forms.
Excess weak acid has been partially neutralised to form the conjugate salt
A skin-care product has a buffered pH of 4.50. The buffer contains a
solution of benzenecarboxylic acid (pKa 4.19, at 25oC) and its salt sodium
benzenecarboxylate. Calculate the proportions of benzenecarboxylic acid (HA) and sodium benzenecarboxylate (A-) in the skin-care product. This is easiest to calculate as the fraction [A-]/[HA]
H+] = 10-pH therefore [H+] = 10-4.50 hence [H+] = 3.16x10-5
Ka = 10-pKa therefore [Ka] = 10-4.19 hence [Ka] = 6.46x10-5
Ka = [H+ ] [A- ]/[HA]
Therefore [A- ]/[HA] = Ka/[H+]
Hence [A-]/[HA] = 6.46x10-5 /3.16x10-5 = 2.04
It could also be expressed as the inverse of this ratio [HA]/[A-] = 0.490
ALWAYS REMEMBER CHARGES SWAP
What is H2CO3 an example of?
natural buffering ion which is responsible for the maintenance of blood pH and also the pH of lakes and streams. It is able to buffer because of this equation: H2CO3 (aq) H+ (aq) + HCO3- (aq)
What happens if we add acid (H+) to H2CO3?
the position of the equilibrium shifts to the left Carbonic acid is formed (which can decompose to make carbon dioxide and water).
What happens if we add an alkali (OH-) to H2CO3?
OH- (aq) this reacts with H+ (aq) and more H2CO3 dissociates,
the equilibrium shifts to the right to restore the H+ (aq) ions.
Healthy blood at a pH of 7.40 has a hydrogencarbonate: carbonic acid ratio of 10.5 : 1.
A patient is admitted to hospital. The patient’s blood pH is measured as 7.20.
Calculate the hydrogencarbonate: carbonic acid ratio in the patient’s blood.
Ka Method Ka = [H][A]/[HA]|
[A-1/[HA] = Ka / [H]
Healthy Blood
[H+] = 10-7.40 = 3.9 x 10-8 mol dm-3 [A-]/[HA] = 10.5 / 1
Ка = (3.9 x 10-8 x 10.5)/1 = 4.18 × 10-7 mol dm-3
Patient’s Blood
[A-1/[HA]
= 4.18 × 10-7/ 10-7.20 = 6.625
= 6.63: 1
What is the enthalpy change of neutralisation?
the energy change when 1 mol
of water is created by the neutralisation of an acid with an alkali under standard
conditions.
What is the end point? (titration curves)
minimum volume of reagent required for the indicator to change
colour. The reagent in the flask has been neutralised completel
What is the equivalence point?
the pH at the centre of the vertical section which occurs at
equimolar portions of acid and base (at the end point)
The pH at the equivalence point as the result of a “competition” between the acid and the base:
• strong acid vs strong base = 7.00
• weak acid vs weak base = 7.00
• strong acid vs weak base < 7.00
• weak acid vs strong base > 7.00
What is an acid base indicator?
weak acid (HA). When in solution, the weak acid has a distinctly different colour to the conjugate base (A−)
What will the indicator appear?
a different colour depending on if it is added to an acidic solution or a basic solution.
What happens when you add H+ to an indicator?
Eqm shifts LHS to use additional H+ ions.
Colour changes towards colour on LHS
What happens when you add OH- to an indicator?
Eqm shifts RHS to use replace H+ ions consumed by OH–
Colour changes towards colour on RHS
