Question

$1.00L$ of a buffer that is $0.100M$ in $HOAc$ ($p{K}_a=4.74)$ and $0.100M$ in $NaOAc$. What is the pH after the addition of $0.0200mol$ $NaOH$? $(log1.5=0.176)$

• A. 4.92
• B. 4.74
• C. 8.74
• D. 6.74

Answer 1

The correct option is A 4.92

Consider the reaction of the strong base with the weak acid first. That reaction goes to completion.
$HOAc+O{H}^{-}\to H_{2}O+OA{c}^{-}$

$\begin{array}{cccc}& \text{HOAc}& O{H}^{-}& OA{c}^{-}\\ \text{Initial}& 100.0mmol& 20.0mmol& 100.0mmol\\ \text{Change}& -20.0mmol& -20.0mmol& +20.0mmol\\ \text{End}& 80.0mmol& 0& 120.0mmol\end{array}$
Using Henderson's equation
$pH=p{K}_a+log\left(\frac{\left[A^{-}\right]}{\left[HA\right]}\right)$
$pH=4.74+log\left(\frac{120.0}{80.0}\right)$
$pH=4.92$