Question

$50\, mL$ of $0.02\, M \,NaOH$ solution is mixed $50\, mL$ of $0.06 \,M$ acetic acid solution, the $pH$ of resulting solution is $\left( p K_{a}\right.$ of acetic acid is $4.76, \log 5=0.70$ )

• A. 5.06
• B. 4.06
• C. 5.46
• D. 4.46

Answer 1

The Correct Option is D

The correct answer is option (D) : 4.46
Given, Volume of \(NaOH\) solution \(=50 \,mL\) 

Molarity of \(NaOH\) solution \(=0.02 \,M\) 

Me of \(NaOH =50 \times 0.02=1 \,meq\) 

Volume of \(CH _{3} COOH\) solution \(=50 \,mL\) 

Molarity of \(CH _{3} COOH =0.06\, M\) 

Me of \(CH _{3} COOH =50 \times 0.06=3\, meq\)
\(1\) me of \(NaOH\) combines with \(3\) me of 

\(CH _{3} COOH\) and forms \(1\) me of \(CH _{3} COONa\). 

\(\therefore 2\) me of \(CH _{3} COOH\) is left. 

So, it forms an acidic buffer. 

Now, for an acidic buffer, 

\(pH = p K_{a}+\log \left[\frac{\text { Salt }}{\text { Acid }}\right]\)
\(pH = p K_{a}+\log \frac{\left[ CH _{3} COONa \right]}{\left[ CH _{3} COOH \right]}\)
\(\therefore pH =4.76+\log \left[\frac{ l }{2}\right]\)
\(pH =4.76+\log \left(5 \times 10^{-1}\right)\)
\(pH =4.76+\log 5-\log 10\)
\(=4.76+0.70- 1\)
\(=4.46\)