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Calculate the amount of `NH_(3)` and `NH_(4)CI` required to prepare a buffer solution of pH `9.0` when total concentration of buffering reagents is `0.6 mol L^(-1)`. `(pK_(b)for NH_(3)=4.7,log 2=0.30)` |
Answer» `pOH=-logK_(b)=log""(["Salt"])/(["Base"])` `5=4.7+log""(a)/(b)` `(a)/(b)=2 :.a=2b` given`a+b=0.6` `2b+b=0.6` `3b=0.6` ,brgt `b=0.2mol e` and `a=0.4mol e` thus `["Salt"]=0.4M`and `[Base]=0.2M` |
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