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 $



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