A) 0.56
B) 1.76
C) 1.18
D) 2.06
E) 1.56
Correct Answer: C
Solution :
For \[pH=1,\,\,[{{H}^{+}}]={{10}^{-1}}M\] For \[pH=2\,[{{H}^{+}}]={{10}^{-2}}M\] \[{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}={{M}_{R}}({{V}_{1}}+{{V}_{2}})\] \[{{10}^{-1}}\times 60+{{10}^{-2}}\times 60={{M}_{R}}(100)\] \[0.6+0.06={{M}_{R}}(100)\] \[{{M}_{R}}=\frac{0.66}{100}=6.6\times {{10}^{-2}}M\] \[{{M}_{R}}=\]Resultant molarity of \[{{H}^{+}}\]ions \[pH=-\log 6.6\times {{10}^{-2}}\] \[=-\,[(\log \,6.6)-2log10]=-\,[0.819-2]\] \[=-\,[-1.18046]=1.18\]
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