A) 3.248 g
B) 4.248 g
C) 1.320 g
D) 6.248 g
Correct Answer: C
Solution :
[c] \[pOH=-\log {{K}_{b}}+\log \frac{[N{{H}_{4}}^{+}]}{[N{{H}_{4}}OH]}\] Let 'a' millimoles of \[N{{H}^{+}}_{4}\]is added to a solution having millimoles of \[N{{H}_{4}}OH=500\times 0.2=100\] \[\therefore [N{{H}_{4}}^{+}]=[salt]=\frac{a}{500}\]and \[[N{{H}_{4}}OH]\] \[=[Base]=\frac{100}{500}\] Given \[{{K}_{b}}\]for \[N{{H}_{4}}OH=2{{\times }^{-5}}\] and \[pH=9\] \[\therefore 5=-\log 2\times {{10}^{-5}}+\log \frac{a/500}{100/500}\] \[\therefore a=200\]millimoles = 0.2 mol Moles of \[{{(N{{H}_{4}})}_{2}}S{{O}_{4}}\]added \[=\frac{a}{2}=0.1\]mol \[\therefore W{{(N{{H}_{4}})}_{2}}S{{O}_{4}}=0.1\times 132=1.32\]
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