Exercise 7.53 – Weak acid calculations - answers

Q753-01 Benzene carboxylic (benzoic) acid has a Ka = 6.6 x 10-5. What is the pH of a 0.30 M aqueous solution of benzene carboxylic acid?

A.  0.52

B.  2.4

C.  4.2

D.  4.7

For the weak acid equilibrium HA H+ + A-

Assuming that the [HA] is the same at equilibrium as in the beginning, and observing that [H+] = [A-] from the equation stoichiometry:
6.6 x 10-5 = [H+][A-]/[0.3]
0.3 x 6.6 x 10-5 = [H+]2
[H+] = 4.45 x 10-3
pH = -log[H+] = 2.35

Q753-02 What is the relationship between Ka and pKa?

A.  pKa = -logKa

B.  pKa = 1.0 x 10-14 Ka

C.  pKa = logKa

D.  pKa = 1.0 Ka

The letter 'p' means the negative log of the following constant, ∴pKa = -log(10)Ka
Response A


Q753-03 If a weak monoprotic acid is 5.0% dissociated in a 0.10 M aqueous solution of the acid, what is the acid equilibrium constant, Ka, for this acid?

A.  2.5 x 10-5

B.  2.5 x 10-4

C.  5.0 x 10-3

D.  5.0 x 10-2

For the weak acid equilibrium HA H+ + A-
5% dissociated means that only 5% of the acids molecules turn to ions (leaving 95% undissociated)
HA / / H+ / + / A-
initially moles / 0.1 / 0 / 0
@ equilibrium / 0.095 / 0.005 / 0.005

Ka = [0.005][0.005]/[0.095]
Ka = 2.63 x 10-4
If we assume the approximation for the IB that the acid concentration is the same at equilibrium then this becomes
Ka = [0.005][0.005]/[0.1]
Ka = 2.5 x 10-4

Q753-04 What are the [H+] and [OH-] in a 0.10 mol dm-3 solution of a weak acid (Ka = 1.0 x 10-3)?

[H+] / [OH-]
A / 1.0 x 10-1 / 1.0 x 10-13
B / 1.0 x 10-3 / 1.0 x 10-11
C / 1.0 x 10-2 / 1.0 x 10-12
D / 1.0 x 10-6 / 1.0 x 10-8
For the weak acid:

Assume [HA] to be unchanged = 0.10 mol dm-3
And [H+] = [A-]
Therefore [H+] = √(Ka x 0.1)
Therefore [H+] = 1.0 x 10-2 mol dm-3
And as [H+] x [OH-] = 1.0 x 10-14
Therefore [OH-] = 1.0 x 10-12 mol dm-3

Q753-05 The Ka value for an acid is 1.0 x 10-2. What is the Kb value for its conjugate base?

A.  1.0 x 10-2

B.  1.0 x 10-6

C.  1.0 x 10-10

D.  1.0 x 10-12

Ka and Kb are connected by the relationship: Ka + Kb = 1.0 x 10-14
Therefore if Ka = 1.0 x 10-2, then Kb = 1.0 x 10-12

Q753-06 What is the Ka of a 0.10 mol dm-3 solution of a weak monoprotic acid if the [H+] = 2.0 x 10-3 mol dm-3?

A.  2.0 x 10-2 mol dm-3

B.  2.0 x 10-4 mol dm-3

C.  4.0 x 10-5 mol dm-3

D.  4.0 x 10-7 mol dm-3

The acid dissociation constant is given by:

And [H+] = [A-]
Therefore this could be written as:

Therefore when [HA] = 0.10 mol dm-3 and [H+] = 2.0 x 10-3 mol dm-3
Ka = 4.0 x 10-5

Q753-07 A 0.1 mol dm-3 solution of a weak acid has a pH of 3.0. What is Ka for this acid?

A.  1 x 10-1

B.  1 x 10-3

C.  1 x 10-5

D.  1 x 10-6

The acid dissociation constant is given by:

The relationship between pH and [H+] is:

When pH = 3 the hydrogen ion concentration = 1 x 10-3 mol dm-3
As [H+] = [A-] we can write the expression as:

∴ Ka = (1 x 10-3)2/0.1 = 1 x 10-5

Q753-08 Find the concentration of an ethanoic acid solution whose pH is measured at 3.5 (Ka = 1.78 x 10-5)

The acid dissociation constant is given by:

The relationship between pH and [H+] is:

When pH = 3.5 the hydrogen ion concentration = 3.16 x 10-4 mol dm-3
As [H+] = [A-] we can write the acid dissociation expression as:

Rearranging:

Therefore the concentration of ethanoic acid = 5.6 x 10-3 mol dm-3

Q753-09 The pH of a solution of vinegar is 3.00. The concentration of OH- ion in this solution is

A.  3.00 M

B.  1 x 10-3 M

C.  1 x 10-11 M

D.  17 M

From the molarity of the acid [H+] = 1 x 10-3
and Kw = [H+][OH-] = 1 x 10-14
Therefore [OH-] = 1 x 10-11 mol dm-3

Q753-10 25cm3 of a 0.1 mol dm-3 solution of hydrochloric acid reacts with 75cm3 of a 0.10 mol dm-3 solution of sodium hydroxide. Calculate the pH of the final solution.

25cm3 of a 0.1 mol dm-3 solution of hydrochloric acid = 0.025 x 0.1 moles = 0.0025 moles HCl
75cm3 of a 0.10 mol dm-3 solution of sodium hydroxide = 0.075 x 0.10 moles = 0.0075 moles NaOH
Equation for the reaction: NaOH + HCl NaCl + H2O
Thus all of the acid is neutralised and 0.0075 - 0.0025 moles of NaOH remain = 0.005 moles NaOH.
Total volume of solution after mixing = 25cm3 + 75cm3 = 100cm3
Therefore molarity of the final solution = 0.005/0.1 mol dm-3 = 0.05 mol dm-3 NaOH solution
Sodium hydroxide is a strong base and completely dissociated so [OH-] = 0.05 mol dm-3
pOH = -log[OH-] = 1.30
pH + pOH = 14
Therefore pH = 14 - 1.30 = 12.7