A) 3.5
B) 2.5
C) 6.5
D) 5.5
Correct Answer: A
Solution :
Given, for HCl, pH = 3. \[[{{H}^{+}}]={{10}^{-3}}={{M}_{1}}.\] \[{{V}_{1}}=2\,ml\] For \[NaOH,\,pH=10,\,[{{H}^{+}}]={{10}^{-10}}\,={{M}_{2}}.\] \[{{V}_{2}}=3\,ml\] \[{{M}_{1}}{{V}_{1}}-{{M}_{2}}{{V}_{2}}={{M}_{R}}({{V}_{1}}+{{V}_{2}})\] \[{{10}^{-3}}\times 2-{{10}^{-10}}\times 3={{M}_{R}}(2+3)\] \[2\times {{10}^{-3}}=5\times {{M}_{R}}\] \[{{M}_{R}}=4\times {{10}^{-4}}=[{{H}^{+}}]\] \[pH=\log (4\times {{10}^{-4}})=4-0.6020=3.39=3.5.\] Hence, the correct option is [a].
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