Question

What will be the $pH$ of the solution ,if 0.01 moles of $HCl$ is dissolved in a buffer solution containing 0.02 moles of propanoic acid ($K=1.34\times {10}^{-5}$) and 0.0152 moles of salt , at ${25}^{0}C$?
[log (0.173) = -0.76]

• A. 3.11
• B. 4.11
• C. 5.11
• D. 6.11

Answer 1

The correct option is B 4.11

The reaction can be written as,

$C{H}_{3}C{H}_{2}CO{O}^{-}+H^{+}\rightleftharpoons C{H}_{3}C{H}_{2}COOH+H_{2}O$

$0.02$ $0.01$

Moles of conjugate base after addition of $HCl=0.0152-0.01=0.0052$

& Moles of propanoic acid after addition of $HCl=0.02+0.01=0.03$

Now $pH=p{K}_a+\log \left[\frac{base}{acid}\right]$

$\Rightarrow p{K}_a=-\log K_a=-\log 1.34\times {10}^{-5}=4.87$

$\therefore pH=4.87+\log \left(\frac{0.0052}{0.03}\right)$

$=4.87+\log \left(0.173\right)$

$=4.87-0.76$

$\Rightarrow pH=4.11$