(1) The document provides the balanced chemical equation for the titration reaction between the weak acid HA and sodium hydroxide. (2) Using the titration data provided, the concentration of HA in Solution A is calculated to be 0.25 M. (3) The molecular weight of HA is calculated to be 60.0 g based on the mass of HA used and its concentration in Solution A. (4) The pKa of HA is calculated to be 4.750 based on the given pH of Solution A and the concentration of HA.

1. Question 1 125 marks] (a) A solution is prepared by dissolving 7.5 g of weak acid HA in water
to give a solution of total volume 500 cm3 (Solution A). It required 50 cm3 of 0.1 M NaOH to
titrate completely 20 cm3 of Solution A (i) Give a balanced equation for the titration reaction.
(ii) Determine the concentration of HA in Solution A. (ii) Determine the molecular weight of
weak acid HA (iv) If the pH of Solution A is 2.676, determine the pKa of HA 2 marks] (v) If
instead, only 25 cm3 of 0.1 M NaOH is added to 20 cm3 of Solution A [1 markl 2 marks] 2
marks] calculate the pH of the solution after reaction. 13 marks
Solution
(i) The balanced chemical equation for the reaction between HA and NaOH is given as
HA (aq) + NaOH (aq) -------> NaA (aq) + H 2 O (l)
(ii) As per the stoichiometric equation,
1 mole NaOH = 1 mole HA
Millimoles of NaOH required = millimoles of HA titrated = (50 cm 3
)*(1 mL/1 cm 3
)*(0.1 M) =
5.0 mmole.
Concentration of HA in Solution A = (millimoles of HA)/(volume of solution in mL) = (5.0
mmole)/[(20 cm 3
)*(1 mL/1 cm 3
)] = 0.25 M (ans).
(iii) The volume of Solution A is 500 cm 3
.
Moles of HA in Solution A = (500 cm 3
)*(1 mL/1 cm 3
)*(1 L/1000 mL)*(0.25 M) = 0.125 mole.
We have taken 7.5 g of the weak acid; therefore,
0.125 mole HA = 7.5 g.
1 mole HA = (1 mole)*(7.5 g/0.125 g) = 60.0 g.
Therefore, molar mass of HA = 60.0 g (ans).

2. (iv) Consider the ionization of HA as below.
HA (aq) --------> H +
(aq) + A -
(aq)
Due to the 1:1 nature of ionization, we must have
[H +
] = [A -
].
Therefore, -log [H +
] = -log [A -
].
We know that –log [H +
] = pH; therefore,
log [A -
] = log [H +
] = -pH.
The acid ionization constant is given as
K a = [H +
][A -
]/[HA]
Take logarithms on both sides,
-log K a = -log [H +
][A -
]/[HA]
====> pK a = -log [H +
] – log [A -
] + log [HA]
====> pK a = pH – log [A -
] + log (0.25 M) (assume ionization of HA is small).
====> pK a = pH – (-pH) + (-0.602) = 2*pH – 0.602
====> pK a = 2*2.676 – 0.602 = 5.352 – 0.602 = 4.750 (ans).