A) 11.30
B) 10.53
C) 2.70
D) 8.35
Correct Answer: A
Solution :
Normality of \[Ca{{(OH)}_{2}}=2\times M=2\times 0.01=0.02\] \[\therefore \] Millimoles of \[Ca{{(OH)}_{2}}=N\times V\] \[=0.02\times 20=0.4\] Normality of \[HCl=1\times M=1\times 0.01=0.01\] \[\therefore \] Millimoles of \[HCl=N\times V=0.01\times 30=0.3\] Excess millimoles of\[Ca{{(OH)}_{2}}=0.4-0.3=0.1\] Total volume of solution \[=20+30=50\,mL\] \[\therefore \] Normality of solution \[=\frac{millimoles}{volume}\] \[=\frac{0.1}{50}=2\times {{10}^{-3}}\] \[ie,\] \[[O{{H}^{-}}]=2\times {{10}^{-3}}\] \[\Rightarrow \] \[pOH=-\log \,(2\times {{10}^{-3}})=2.7\] \[\therefore \] \[pH=14-pOH\] \[\therefore \] \[pH=14-2.7=11.3\]
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