Question

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