A) \[N{{H}_{3}}\] of \[0.01\,\,M\,\,HCl\]
B) \[_{10}\frac{1}{{{K}_{a}}}\] of \[0.01\,\,M\,\,{{H}_{2}}S{{O}_{4}}\]
C) \[50\,\,ml\] of \[_{10}\frac{1}{{{K}_{a}}}\]
D) Mixture of 50 \[ml\] of \[0.02\,\,M\,\,{{H}_{2}}S{{O}_{4}}\] and \[50\,\,ml\] of \[0.02\,\,M\,\,NaOH\]
Correct Answer: A
Solution :
M.eq. of 0.01 \[M\,\,HCl=\frac{.01\times 100}{1000}=1\times {{10}^{-3}}\] \[pH=3\] M.eq. of .02 \[M\,\,{{H}_{2}}S{{O}_{4}}=\frac{.04\times 50}{1000}=2\times {{10}^{-3}}\] M.eq. of .02\[M\,\,NaOH=\frac{0.02\times 50}{1000}=1\times {{10}^{-3}}\] Left \[[{{H}^{+}}]=2\times {{10}^{-3}}-1\times {{10}^{-3}}=1\times {{10}^{-3}}\];\[pH=3\]
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