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InterviewSolution
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What will be the amount of `(NH_(4))_(2)SO_(4)` (in g) which must be added to 500 mL of 0.2 M `NH_(4)OH` to yield a solution of pH 9.35? `["Given," pK_(a)"of "NH_(4)^(+)=9.26,pK_(b)NH_(4)OH=14-pK_(a)(NH_(4)^(+))]`A. 5.35B. 6.47C. 10.03D. 7.34 |
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Answer» Correct Answer - A `pK_(a)"of"NH_(4)^(+)=9.26` Hence, `pK_(b)"of"NH_(4)OH=14-9.26=4.74` `pOH=14-9.35=4.65` Each 1 mol of `(NH_(4))_(2)SO_(4)` furnishes 2 moles of `NH_(4)^(+)` in solution. Let `[(NH_(4))_(2)SO_(4)]=x"mol L"^(-1)` `[NH_(4)^(+)]"2 x mol L"^(-1)` `[NH_(4)OH]=0.2"mol L"^(-1)` using Henderson - Hasselbalc equation, `pOH=pK_(b)+"log"([NH_(4)^(+)])/([NH_(4)OH])` `4.65=4.74+log((2x)/(0.2))` This gives `x=0.081"mol L"^(-1)` `=0.0405` mol in 500 mL (as required) `=0.0405xx132=5.35` g |
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