Equilibrium Question 327
Question: What is the pH of a 1M $ CH_3COOH $ a solution $ K_{a} $ of acetic acid $ =1.8\times {10^{-5}}. $
$ K={10^{-14}}mol^{2}litr{e^{-2}} $ [DPMT 2002]
Options:
A) 9.4
B) 4.8
C) 3.6
D) 2.4
Show Answer
Answer:
Correct Answer: A
Solution:
$ CH_3CO{O^{-}}+H_2O $ ⇌ $ CH_3COOH+O{H^{-}} $ \ $ [O{H^{-}}]=c\times h;h=\sqrt{\frac{K_{w}}{K_{a}}\times c} $
$ =\sqrt{\frac{{10^{-14}}}{1.8\times {10^{-5}}}\times 1} $
$ =2.35\times {10^{-5}} $ \ $ pOH=4.62 $ ; $ pH=9.38\approx 9.4 $
