Question

What will be the $ pH $ of solution formed by mixing $ 10 \,mL 0.1 \,M $ $ NaH_{2}PO_{4} $ and $ 15 \,mL $ $ 0.1 \,M Na_{2}HPO_{4} $ . [Given: $ Pk_{1}=2.12, pK_{2} =7.2 $ ]

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

Answer 1

Answer: (C)

Given, (For : $NaH_{2}PO_{4})$
$10 \,mL$ of $0.1 \,M \,NaH_{2}PO_{4}$ and
$PK_{1}=2.12$
$\because PH_{1}=PK_{1}+log \frac{\left[A^{-}\right]}{\left[H_{A}\right]}=\frac{\left[0.1\times2\right]}{1}$
Here, $\left[H_{A}\right]=\left[NaH_{2}PO_{4}\right]=0.1\times10=1$
$\left[A^{-}\right]=\left[H^{+}\right]$
$\therefore PH_{1}=2.12+log\left[\frac{0.2}{1}\right]$
$PH_{1}=2.12+\left(-0.6989\right)$
$PH_{1}=1.421$
For $Na_{2}HPO_{4}$,
$15\, mL$ of $0.1 M Na_{2}HPO_{4}$ and $PK_{2}=7.2$
$\because PH_{2}=PK_{2}+log\left[\frac{A^{-}}{HA}\right]$
Here $\left[HA\right]=\left[NaHPO_{4}\right]$
$\left[A^{-}\right]=\left[H^{+}\right]$
$PH_{2}=7.2+log \left[\frac{0.1}{1.5}\right]$
$PH_{2}=7.2+\left(-1.1739\right)$
$pH_{2}=6.0261$
Total $pH =PH_{1}+PH_{2}$
$=1.421+6.0261=7.4471$