Question

To produce a buffer solution that has a pH of 5.270 with already existing solution that contains 10.0 mmol of acetic acid in 1L of solution. How many millimoles of sodium acetate to added to this solutions?
When $p{K}_a$ of acetic acid is 4.752.
$log32.96=1.518$

• A. 25
• B. 10
• C. 23
• D. 33

Answer 1

The correct option is D 33

Given, $p{K}_a=4.752,pH=5.27,V=1L,\text{mole of acetic acid}=10.0mmol$
Using the Henderson-Hasselbalch equation for weak acid:
$pH=p{K}_a+log\left[\frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}\right]$
$pH=4.752+log\left[\frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}\right]$
$5.27=4.752+log\left[\frac{\text{number of mole of sodium acetate}}{\text{number of mole of acetic acid}}\right](\because V=1L)$
$5.27=4.752+log\left[\frac{\text{number of mole of sodium acetate}}{10.0}\right]$
$log\left[\frac{\text{number of mole of sodium acetate}}{10.0}\right]=0.518$
$log\left(\text{number of mole of sodium acetate}\right)-log10.0=0.518$
$log\left(\text{number of mole of sodium acetate}\right)=1.518(\because log10=1)$
Number of mole of sodium acetate$=32.96mmol$