Equilibrium Question 410
Question: What will be the $ pH $ of a $ {10^{-8}},M,HCl $ solution [MP PET/PMT 1998; RPET 1999;MP PMT 2000]
Options:
A) 8.0
B) 7.0
C) 6.98
D) 14.0
Show Answer
Answer:
Correct Answer: C
Solution:
$ H_2O $ ⇌ $ [{H^{+}}][O{H^{-}}] $ HCl ⇌ $ [{H^{+}}][C{l^{-}}] $
Total $ [{H^{+}}]={{[{H^{+}}]} _{H_2O}}+{{[{H^{+}}] } _{HCl}} $
$ ={10^{-7}}+{10^{-8}} $
$ ={10^{-7}}[1+{10^{-1}}] $
$ [{H^{+}}]={10^{-7}}\times \frac{11}{10} $
$ $
$ pH=-\log ,[{H^{+}}]=-\log ( {10^{-7}}+\frac{11}{10} ) $ ; $ pH=6.958 $
