Question

Calculate the pH of a buffer solution made from 0.20 $MH{C}_2{H}_3{O}_2$ and 0.50 M $C_2{H}_3{O}_2^{-}$ that has an acid dissociation constant for $H{C}_2{H}_3{O}_2$ of $1.8\times {10}^{-5}.$
log(1.8 )=0.255, log(2.5 )=0.397

• A. 7
• B. 5.14
• C. 5.64
• D. 4.35

Answer 1

The correct option is B 5.14

Given, $\left[C_2{H}_3{O}_2^{-}\right]=0.50M,\left[H{C}_2{H}_3{O}_2\right]=0.2M,K_a=1.8\times {10}^{-5}.$
Using the Henderson-Hasselbalch equation for weak acid:
$pH=p{K}_a+\log \left[\frac{\left[\text{conjugate base}\right]}{\left[acid\right]}\right]$
$pH=p{K}_a+\log \left(\frac{\left[A^{-}\right]}{\left[HA\right]}\right)$
$pH=p{K}_a+\log \left(\frac{\left[C_2{H}_3{O}_2^{-}\right]}{\left[H{C}_2{H}_3{O}_2\right]}\right)$
$pH=-\log (1.8\times {10}^{-5})+log\left(\frac{0.50}{0.20}\right)$
$pH=-\log (1.8\times {10}^{-5})+log\left(2.5\right)$
$pH=5-\log \left(1.8\right)+log\left(2.5\right)$
$pH=5-0.255+0.397$
$pH=5.14$