Question

A buffer solution, 0.080 M in $N{a_2}P{O_4}$ and 0.020 M in $N{a_3}P{O_4}$ is prepared. The electrolytic oxidation of 1.0 mmol of the organic compound $RNHOH$ is carried out in 100ml of the
buffer. The reaction is as follows:
$RNHOH + {H_2}O \to RN{O_2} + 4{H^ + } + 4{e^ - }$
Calculate the approximate pH of the solution after the oxidation is complete.
(A) 6.19
(B) 7.81
(C) 10.34
(D) 12.45

Answer 1

Buffer solution is the mixture of weak acid or weak base, along with conjugate acid and conjugate base.
pH is potenz hydrogen, power of hydrogen and it shows acidic and basic strength of solution. It depends
on ${H^ + }$ ion concentration. There is a direct formula for calculating pH of the buffer solution.
$pH = p{K_a} + \log \dfrac{{[salt]}}{{[acid]}}$

Complete step by step answer:
Buffer solution is a special type of solution, in which when we add little amount of acid or base, still the
the pH value of the solution does not change.
From the reaction below we can find out concentration of ${H^ + }$ ions
$RNHOH + {H_2}O \to RN{O_2} + 4{H^ + } + 4{e^ - }$

1 mmol of organic compound will produce 4 mmol of ${H^ + }$ ions.
To find out concentration of ${H^ + }$ ions, we need moles and volume of solution, volume is given as
100 ml.
$\therefore [{H^ + }] = \dfrac{{moles}}{{Volume}}$
Moles = 0.004
Volume = 0.1 litre
Substitute the values, we get:
$[{H^ + }] = \dfrac{{0.004}}{{0.1}}$
$\therefore [{H^ + }] = 0.04M$
We know another reaction:
$P{O_4}^{ - 3} + {H^ + } \to HP{O_4}^{2 - }{\text{ }}\dfrac{1}{{{k_3}}}$

$$0.02{\text{ 0}}{\text{.04 0}}{\text{.08}} \\\ \- {\text{ 0}}{\text{.02 0}}{\text{.1}} \\\$$

Another reaction can be written as:
$HP{O_4}^{2 - } + {H^ + } \to {H_2}P{O_4}^ - {\text{ }}\dfrac{1}{{{k_2}}}$

$$0.1{\text{ 0}}{\text{.02 }} - \\\ 0.08{\text{ }} - {\text{ 0}}{\text{.02}} \\\$$

From given data, ${k_2} = 6.3 \times {10^{ - 8}}$
We know to find out pH of buffer, the formula is:
$pH = p{K_a} + \log \dfrac{{[salt]}}{{[acid]}}$
Substituting values in above equation we get:

$pH = p{k_2} + \log \dfrac{{[N{a_2}P{O_4}]}}{{[{H_2}P{O_4}^ - ]}}$
Substitute the numerical values into above equation:
$pH = 7.2 + \log \dfrac{{0.08}}{{0.02}}$
On dividing the value 0.08 with 0.02, we get
$pH = 7.2 + \log 4$
We know from log table, the value of log 4 = 0.6,
$pH = 7.2 + 0.6$
On simplification, we get:
$\therefore pH = 7.8$
Thus the correct option is (B) 7.81.

Note:
Take care of values of equilibrium constant, especially for weak acids used in question. Also
logarithmic values can be remembered. There are 2 types of buffer solution- acidic buffer and basic
buffer, so we should be careful in making reactions, buffer solution and calculations using correct formula for the acidic buffer in the above question.