Question

Two buffers, $\left(X\right)$ and $\left(Y\right)$ of $pH$ 4.0 and 6.0 respectively are prepared from acid HA and the salt $NaA$. Both the buffers are $0.50M$ of HA. What would be the pH of the solution obtained by mixing equal volumes of the buffers ? $K_{HA}=1.0\times {10}^{-5}$

• A. 8.4
• B. 4.3
• C. 5.7
• D. 7.5

Answer 1

The correct option is C 5.7

Henderson equation is given by $pH=-{\text{log}}_{10}Ka+{\text{log}}_{10}\frac{\left[Salt\right]}{\left[Acid\right]}$
When $pH=4$
$4=-{\text{log}}_{10}1.0\times {10}^{-5}+{\text{log}}_{10}\frac{\left[\text{Salt}\right]}{0.5}$
$4=5+{\text{log}}_{10}\frac{\left[\text{Salt}\right]}{0.5}$
${\text{log}}_{10}\frac{\left[\text{Salt}\right]}{0.5}=-1$
$\left[\text{Salt}\right]=0.1\times 0.5=0.05$
When $pH=6$,
$6=-{\text{log}}_{10}(1.0\times {10}^{-5})+{\text{log}}_{10}\frac{\left[\text{Salt}\right]}{0.5}$
$6=5+{\text{log}}_{10}\frac{\left[\text{Salt}\right]}{0.5}$
${\text{log}}_{10}\frac{\left[\text{Salt}\right]}{0.5}=1$
$\left[\text{Salt}\right]=10\times 0.5=5M$
Suppose $V$ litre each of both buffers are mixed. The concentration of the salt in this solution will be:
$\left[\text{Salt}\right]=\frac{(0.05\times V)+(5\times V)}{2V}=\frac{5.05}{2}M$
Concentration of $\left[HA\right]$ in mixed buffer
$=\frac{(0.5\times V)+(0.5\times V)}{2V}=0.5M$
$pH=-{\text{log}}_{10}(1.0\times {10}^{-5})+{\text{log}}_{10}\frac{\left[\frac{5.05}{2}\right]}{0.5}$
$=5+0.7033=5.7033$