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`pH` of a mixture containing `0.10 M X^(-)` and `0.20 M HX` is: `[pK_(b)(X^(-))=4]`A. `4+log2`B. `4-log2`C. `10+log2`D. `10-log2` |
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Answer» `pOH=pK_(b)+log((["Salt"])/(["Base"]))` Let a mol litre^(-1) be concentration of salt., then concentration of base `=(0.29-a)mol//L`. `4.4=-log1.8xx10^(-5)+log((a)/((0.29-a)))` `:.a=0.09` `["Salt"]=0.9M` & `["Base"]=0.29-0.09=0.20M` |
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