15 Acid-Base Equilibria Flashcards
Which of the following is correct when separate 20.0 cm3 samples of HCl(aq) and CH3COOH(aq) of equal concentrations is titrated with 1.0 mol dm-3 NaOH(aq)?
A) Both HCl(aq) and CH3COOH(aq) require the same volume of NaOH(aq) for titration.
B) Compared to CH3COOH(aq), HCl(aq) requires a greater volume of NaOH(aq) for titration.
C) Compared to HCl(aq), CH3COOH(aq) requires a greater volume of NaOH(aq) for titration.
A
True or False: HCl is a stronger acid than CH3COOH, thus producing a higher concentration of H+ ions that requires more NaOH for a complete reaction.
False. CH3COOH(aq) ⇌ CH3COO–(aq) + H+(aq) - (1)
The addition of NaOH causes the equilibrium position for the dissociation of CH3COOH to shift right until dissociation is complete. Hence, OH- continuously depletes H+ in titration, forming H2O. The fall in [H+] causes equilibrium position (1) to shift right until dissociation is complete and CH3COOH reacts completely with NaOH.
Since the amount of CH3COOH equals the amount of HCl, the amount (hence volume) of NaOH required for titration will be the same.
True or False: CH3COOH is a weaker acid than HCl, thus reacting more slowly with NaOH.
False. The rate of the neutralisation reaction will not affect the volume of titrant needed.
When 25 cm3 of 0.100 mol dm–3 ethanoic acid, CH3COOH, is titrated with 0.100 mol dm–3 NaOH(aq), what is the volume of NaOH(aq) required for the titration?
Since CH3COOH reacts with NaOH in a 1 : 1 ratio, and the concentrations of CH3COOH and NaOH are the same, 25 cm3 of CH3COOH will require 25 cm3 of NaOH for titration.
When 25 cm3 of 0.100 mol dm–3 ethanoic acid, CH3COOH, is titrated with 0.100 mol dm–3 NaOH(aq), what is the pH of the solution when half the volume of NaOH (aq) required for titration is added?
(Ka of CH3COOH is 1.8 x 10–5 mol dm–3 at 298 K)
At V cm3 of NaOH added, half the CH3COOH has been neutralised to form the salt, CH3COONa. At this point, [CH3COOH] = [CH3COO–] and the solution is a buffer at maximum buffering capacity.
Hence, pH = pKa = -lg (1.8 x 10–5) = 4.74
Before the equivalence point is reached), some of the CH3COOH has been neutralised to form the salt, CH3COONa. Since the solution contains a mixture of the weak acid, CH3COOH, and its conjugate base, CH3COO–, the solution is a buffer solution.
When 25 cm3 of 0.100 mol dm–3 ethanoic acid, CH3COOH, is titrated with 0.100 mol dm–3 NaOH(aq), is the pH at the equivalence point smaller than/equal to/greater than 7?
At the equivalence point, all the CH3COOH has been neutralised to form the salt, CH3COONa. Since CH3COO– is the conjugate base of the weak acid CH3COOH, CH3COO– can undergo hydrolysis to form OH- ions, resulting in pH to be greater than 7.
CH3COO–(aq) + H2O(l) ⇌ CH3COOH(aq) + OH–(aq)
What is an Arrhenius Acid?
An acid that releases H+ ions in an aqueous solution.
What is a Brønsted-Lowry Base
A proton acceptor
What is a Lewis Base
An electron-pair donor
Which of the following is a conjugate acid-base pair?
A) H2SO4 and HSO4-
B) H3PO3 and H3PO4
C) H2CO3 and CO32-
D) C6H5NH3+ and C6H5NH2
In a conjugate acid-base pair, the acid and base differ from each other by one proton, H+. Hence, only options A and D show a conjugate acid-base pair.
The pH of solution A is 10 while that of solution B is 12.5. Which of the following conclusions can be made about the two solutions?
A) A and B contain a monoacidic base.
B) B is a stronger base than A.
C) Solution B has a higher concentration of OH- ions than solution A.
D) B has a higher concentration than A
C) Solution B has a higher concentration of OH- ions than solution A.
The pH of normal human blood is 7.4. Strenuous exercise can cause the condition called acidosis in which the pH falls. If the pH drops to 6.8, death may occur.
How many times greater is the hydrogen ion concentration in blood at pH 6.8 compared with that at pH 7.4?
At pH = 7.4, [H+] = 10-pH
= 10-7.4
=3.98 x 10-8 mol dm-3
At pH = 6.8, [H+] = 10-pH
= 10-6.8
=1.58 x 10-7 mol dm-3
Therefore, [H+] at pH 6.8 is (1.58 x 10-7) (3.98 x 10-8) = 4.0 times greater than at pH 7.4
Like H2O, H2PO4– can behave as both an acid and a base. You are given the following Ka and Kb values of H2PO4–.
Ka of H2PO4– = 6.20 x 10–8 mol dm−3
Kb of H2PO4– = 1.33 x 10–12 mol dm–3
Determine the Ka of H3PO4 and Kb of HPO42–.
Using the concept of conjugate acid-base pair, identify that
H3PO4 (acid) and H2PO4–(base) is a conjugate acid-base pair
H2PO4– (acid) and HPO42–(base) is a conjugate acid-base pair
Applying the formula of Kw = Ka x Kb for a conjugate acid-base pair,
Ka of H3PO4 = Kw Kb of H2PO4– = 7.52 x 10–3 mol dm−3
Kb of HPO42– = Kw Ka of H2PO4– = 1.61 x 10–7 mol dm-3
Which of the following could act as buffer solutions?
1) H2CO3 ad NaHCO3
2) CH3NH2 and CH3NH3Cl
3) HCl and NaCl
A buffer consists of a mixture of a weak acid and its conjugate base / weak base and its conjugate acid.
Option 1 can act as a buffer as it contains the weak acid, H2CO3, and its conjugate base, HCO3–.
Option 2 can act as a buffer as it contains the weak base, CH3NH2, and its conjugate acid, CH3NH3+.
Option 3 cannot act as a buffer as it contains only the strong acid, HCl, and its conjugate base, Cl–.
Calculate the pH of a solution containing 0.20 mol dm–3 of benzoic acid, C6H5COOH, and 0.5 mol dm–3 of potassium benzoate, C6H5COO–K+.
(Ka of benzoic acid, C6H5COOH = 6.5 x 10−5 mol dm−3)
Applying the Henderson-Hasselbalch equation for an acidic buffer:
pH = pKa + lg ([A-]/[HA]) = -lg Ka + lg ([C6H5COO-]/[C6H5COOH]) = -lg (6.5 x 10−5) + lg (0.5/0.2) = 4.59
