A) \[2\]
B) \[1.3\]
C) \[0\]
D) \[7\]
Correct Answer: B
Solution :
We know that, \[20ml.\]of \[0.1N\text{ }HCl\] contains \[[{{H}^{+}}]=\frac{20}{1000}\times 0.1=2\times {{10}^{-3}}moles.\] Similarly \[20ml\]. of \[.001\text{ }N\text{ }KOH\] contains \[[O{{H}^{-}}]=\frac{20}{100}\times 0.001=2\times {{10}^{-5}}moles\] The \[[{{H}^{+}}]\] in the mixture \[=(2\times {{10}^{-3}})(2\times {{10}^{-5}})=2\times 99\times {{10}^{-5}}\] \[1000\text{ }ml\]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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