Question

$50mL$ of $0.1MHCl$ and $50mL$ of $0.2MNaOH$ are mixed. The $pH$ of the resulting solution is

• A. 1.30
• B. 4.2
• C. 12.70
• D. 11.70

Answer 1

Answer: (C)

$50ML$ of $0.1MHCl=\frac{0.1\times 50}{1000}=5\times 10^{-3}50ML$ of
$0.2MNaOH=\frac{0.2\times 50}{100}=10\times 10^{-3}$
Hence, after neutralisation $NaOH$ is left.
$=10\times 10^{-3}-5\times 10^{-3}=5\times 10^{-3}$
Total volume $=100cc$ The concentration of
$NaOH=\frac{5\times 10^{-3}\times 1000}{100}=0.05M$
$\left[O{H}^{-}\right]=0.05M=5\times 10^{-2}M$
$pOH=-\log \left[O{H}^{-}\right]=-\log \left[5\times 10^{-2}\right]$
$=1.3010pH+pOH=14$
$pH=14-1.3010=12.699=12.70$