Question

To a solution of acetic acid, solid sodium acetate is added gradually. When x mole of salt is added, the pH has a certain value. If y mole of salt is added, the pH change by $0.6$ units to the previous pH. What is the ratio of x and y?

Answer 1

$pH$ of a buffer solution of weak acid and its salt-

$pH=p{K}_a+\log \frac{\left[\text{salt}\right]}{\left[\text{acid}\right]}$

Let the volume of solution is $V$.

Therefore,

When $x$ mole of salt is added-

Concentration of salt will be $\frac{x}{V}$.

$pH=p{K}_a+\log \frac{\left(\frac{x}{V}\right)}{\left[\text{acid}\right]}.....\left(1\right)$

When $y$ mole of salt is added pH is changed by 0.6-

Concentration of salt will be $\frac{y}{V}$.

$pH+0.6=p{K}_a+\log \frac{\left(\frac{y}{V}\right)}{\left[\text{acid}\right]}.....\left(2\right)$

Subtracting equation $\left(1\right)$ from $\left(2\right)$, we have

$(pH+0.6)-pH=(p{K}_a+\log \frac{\left(\frac{y}{V}\right)}{\left[\text{acid}\right]})-(p{K}_a+\log \frac{\left(\frac{x}{V}\right)}{\left[\text{acid}\right]})$

$\Rightarrow 0.6=p{K}_a+\log \frac{\left(\frac{y}{V}\right)}{\left[\text{acid}\right]}-p{K}_a-\log \frac{\left(\frac{x}{V}\right)}{\left[\text{acid}\right]}$

$\Rightarrow 0.6=\log \frac{y}{x}$

$\Rightarrow \frac{y}{x}={10}^{0.6}\approx 4$

$\therefore \frac{x}{y}=\frac{1}{4}$

Hence $x:y=1:4$.