A) 1.30
B) 4.2
C) 12.70
D) 11.70
Correct Answer: C
Solution :
50 mL of\[0.1M\,HCl=\frac{0.1\times 50}{1000}=5\times {{10}^{-3}}\] 50 mL of \[0.2\text{ }M\text{ }NaOH=\frac{0.2\times 50}{1000}=10\times {{10}^{-3}}\] Hence, after neutralisation\[NaOH\]left \[=10\times {{10}^{-3}}-5\times {{10}^{-3}}=5\times {{10}^{-3}}\] Total volume\[=100\text{ }cc\] The concentration of\[NaOH\] \[=\frac{5\times {{10}^{-3}}\times 1000}{100}=0.05\,M\] \[{{[OH]}^{-}}=0.05\,M=5\times {{10}^{-2}}M\] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log [5\times {{10}^{-2}}]\] \[=1.3010\] \[pH+pOH=14\] \[pH=14-1.3010\] \[=12.699=12.70\]
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