Equilibrium Question 792
Question: A solution of $ 0.1MNaZ $ has $ pH=8.90. $ The $ K_{a} $ of HZ is
Options:
A) $ 6.3\times {10^{-11}} $
B) $ 6.3\times {10^{-10}} $
C) $ 1.6\times {10^{-5}} $
D) $ 1.6\times {10^{-6}} $
Show Answer
Answer:
Correct Answer: C
Solution:
NaZ is salt of $ W_{A}/S_{B} $
$ \therefore pH=\frac{1}{2}(pK_{W}+pK_{a}+logC) $
$ 8.9\times 2=14+pK_{a}+log0.1 $
$ 17.8=14+pK_{a}-1 $
$ pK_{a}=4.8, $
$ K_{a} $ = Antilog $ ( -4.8 ) $
$ =1.585\times {10^{-5}}\approx 1.6\times {10^{-5}} $
