A) \[6.3\times {{10}^{-11}}\]
B) \[6.3\times {{10}^{-10}}\]
C) \[1.6\times {{10}^{-5}}\]
D) \[1.6\times {{10}^{-6}}\]
Correct Answer: C
Solution :
[c] NaZ is salt of \[{{W}_{A}}/{{S}_{B}}\] \[\therefore pH=\frac{1}{2}(p{{K}_{W}}+p{{K}_{a}}+logC)\] \[8.9\times 2=14+p{{K}_{a}}+log0.1\] \[17.8=14+p{{K}_{a}}-1\] \[p{{K}_{a}}=4.8,\] \[{{K}_{a}}\]= Antilog \[\left( -4.8 \right)\] \[=1.585\times {{10}^{-5}}\approx 1.6\times {{10}^{-5}}\]
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