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The `pH` of `0.02 M NH_(4)Cl (aq) (pK_(b)=4.73)` is equal toA. `3.78`B. `4.73`C. `5.48`D. `7.00` |
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Answer» Correct Answer - C `pH=(1)/(2)[pK_(w)-pK_(b)-log C]` `= (1)/(2)[14-4.73-log 0.02]` `= (1)/(2)[14-4.73+1.698]=5.48` |
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