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In a buffer solution containing equal concentration of `B^(-)` and `HB`, the `K_(b)` for `B^(-)` is `10^(-10)`. The `pH` of buffer solution isA. 10B. 7C. 6D. 4 |
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Answer» Correct Answer - D Key Idea - (i) For basic buffer, `pOH=pK_(b)+log.(["salt"])/(["base"])` (ii) `pH+pOH = 14` Given, `K_(b)=1xx10^(-10)`, [salt]=[base] `pOH=-log K_(b)+log.(["salt"])/(["base"])` `therefore pOH=-log(1xx10^(-10))+log1=10` `pH+pOH=14 [because "concentration of "[B^(-)]=[HB]` `pH=14-10=4` |
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