Question

What will be the $pH$ of solution formed by mixing $10mL0.1M$ $Na{H}_{2}P{O}_4$ and $15mL$ $0.1MN{a}_{2}HP{O}_4$ . [Given: $P{k}_1=2.12,p{K}_2=7.2$ ]

• A. $7.0$
• B. $6.9$
• C. $7.4$
• D. $7.5$

Answer 1

Answer: (C)

Given, (For : $Na{H}_{2}P{O}_4)$
$10mL$ of $0.1MNa{H}_{2}P{O}_4$ and
$P{K}_1=2.12$
$\because P{H}_1=P{K}_1+log\frac{\left[A^{-}\right]}{\left[H_A\right]}=\frac{\left[0.1\times 2\right]}{1}$
Here, $\left[H_A\right]=\left[Na{H}_{2}P{O}_4\right]=0.1\times 10=1$
$\left[A^{-}\right]=\left[H^{+}\right]$
$\therefore P{H}_1=2.12+log\left[\frac{0.2}{1}\right]$
$P{H}_1=2.12+\left(-0.6989\right)$
$P{H}_1=1.421$
For $N{a}_{2}HP{O}_4$,
$15mL$ of $0.1MN{a}_{2}HP{O}_4$ and $P{K}_2=7.2$
$\because P{H}_2=P{K}_2+log\left[\frac{A^{-}}{HA}\right]$
Here $\left[HA\right]=\left[NaHP{O}_4\right]$
$\left[A^{-}\right]=\left[H^{+}\right]$
$P{H}_2=7.2+log\left[\frac{0.1}{1.5}\right]$
$P{H}_2=7.2+\left(-1.1739\right)$
$p{H}_2=6.0261$
Total $pH=P{H}_1+P{H}_2$
$=1.421+6.0261=7.4471$