question_answer

20 mL of 0.2 M \[NaOH\] are added to 50 mL of 0.2 M \[C{{H}_{2}}COOH\]\[({{K}_{a}}=1.8\times 10{{~}^{-5}})\] the pH of the solution is

A) 4.56                 

B) 4.73  

C) 9.45                 

D) 6.78

Correct Answer: A

Solution :

[a]  \[Conc.=\frac{\text{Millimoles}}{\text{Total}\,\text{Volume}}\] \[[C{{H}_{3}}COOH]=\frac{6}{70};[C{{H}_{3}}COONa]=\frac{4}{70}\] \[pH=-log1.8\times {{10}^{-5}}+log\frac{4/70}{6/70},pH=4.56\]