Question

A chemistry graduate student is given 125.mL of a 0.20M acetic acid HCH3CO2 solution. Acetic acid is a weak acid with =Ka×1.810−5. What mass of NaCH3CO2 should the student dissolve in the HCH3CO2 solution to turn it into a buffer with pH =4.47?

Answer 1

Answer : The mass of sodium acetate is, 1.097 grams.

Explanation : Given,

The dissociation constant for acetic acid = $K_a=1.8\times 10^{-5}$

Concentration of acetic acid (weak acid)= 0.20 M

volume of solution = 125. mL

pH = 4.47

First we have to calculate the value of $pK_a$.

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (1.8\times 10^{-5})$

$pK_a=5-\log (1.8)$

$pK_a=4.74$

Now we have to calculate the concentration of sodium acetate (conjugate base or salt).

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$4.47=4.74+\log (\frac{[Salt]}{0.20})$

$[Salt]=0.107M$

Now we have to calculate the mass of sodium acetate.

$\text{Concentration}=\frac{\text{Mass of }NaCH_3CO_2\times 1000}{\text{Molar mass of }NaCH_3CO_2\times \text{Volume of solution (in mL)}}$

$0.107M=\frac{\text{Mass of }NaCH_3CO_2\times 1000}{82g/mol\times 125mL}$

$\text{Mass of }NaCH_3CO_2=1.097g$

Therefore, the mass of sodium acetate is, 1.097 grams.