A) 1.30
B) 4.2
C) 12.70
D) 11.70
Correct Answer: C
Solution :
50 ML of \[0.1\,M\,HCl=\frac{0.1\times 50}{1000}=5\times {{10}^{-3}}\] 50 ML of \[0.2\,M\,NaOH=\frac{0.2\times 50}{100}=10\times {{10}^{-3}}\] Hence, after neutralisation \[\text{NaOH}\]is left. \[\text{=10}\times \text{1}{{\text{0}}^{-3}}-5\times {{10}^{-3}}\] \[=5\times {{10}^{-3}}\] Total volume = 100 cc The concentration of \[\text{NaOH}\] \[=\frac{5\times {{10}^{-3}}\times 1000}{100}=0.05\,M\] \[[O{{H}^{-}}]=0.05\,M=5\times {{10}^{-2}}M\] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log [5\times {{10}^{-2}}]\] \[=1.3010\] \[pH+pOH=14\] \[pH=14-1.3010\] \[=12.699=12.70\]
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