A) 2
B) 1.3
C) 0
D) 7
Correct Answer: B
Solution :
We know that, 20 ml. of \[0.1\,N\,HCl\]contains \[[{{H}^{+}}]=\frac{20}{1000}\times 0.1=2\times {{10}^{-3}}\]moles. Similarly 20 ml. of 0.001 N KOH contains \[[O{{H}^{-}}]=\frac{20}{100}\times 0.001=2\times {{10}^{-5}}\]moles The \[[{{H}^{+}}]\] in the mixture \[=(2\times {{10}^{-3}})(2\times {{10}^{-5}})=2\times 99\times {{10}^{-5}}\] 1000 ml of mixture \[[{{H}^{+}}]=\frac{1000\times 2\times 99\times {{10}^{-5}}}{40}=4950\times {{10}^{-5}}\] \[\because \]\[pH=-\log [{{H}^{+}}]=5-log\,4950\] \[=5-3.6946\,=1.3\]
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