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The pH of a 0.1 M solution of `NH_(4)OH` (having `K_(b)=1.0xx10^(-5))` is equal toA. 10B. 6C. 11D. 12 |
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Answer» Correct Answer - C `[OH^(-)]=sqrt(K_(b)xxC)` `=sqrt(1xx10^(-5)xx10^(-1))` `sqrt(10^(-6))=10^(3)` `K_(w)=[H^(+)][OH^(-)]` `10^(-14)=[H^(+)][OH^(-3)]` `[H^(+)]=10^(-11)` Hence, `pH=- logH^(+)` `=-log(1xx10^(11))=11` |
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