Question

A chemistry graduate student is given 125. mL of a 1.30 M propanoic acid (HC2H,Co2) solution. Propanoic acd is a weak acid with K - 1.3 10-5. what mass of KC2H CO2 should the student dissolve in the HC2H,CO2 solution to turn it into a buffer with pH 5.02?

Answer 1

Answer : The  mass of $KC_2H_5CO_2$ is, 24.5 grams

Explanation : Given,

$K_a=1.3\times 10^{-5}$

pH = 5.02

Concentration of $HC_2H_5CO_2$ = 1.30 M

Volume of solution = 125 mL = 0.125 L

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:

$pK_a=-\log (1.3\times 10^{-5})$

$pK_a=5-\log (1.3)$

$pK_a=4.89$

Now we have to calculate the concentration of $KC_2H_5CO_2$

Using Henderson Hesselbach equation :

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

$pH=pK_a+\log \frac{[KC_2H_5CO_2]}{[HC_2H_5CO_2]}$

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

$5.02=4.89+\log (\frac{[KC_2H_5CO_2]}{1.30})$

$[KC_2H_5CO_2]=1.75M$

Now we have to calculate the moles of $KC_2H_5CO_2$

$\text{Moles of }KC_2H_5CO_2=1.75M\times 0.125L=0.219mol$

Now we have to calculate the mass of $KC_2H_5CO_2$

$\text{Mass of }KC_2H_5CO_2=\text{Moles of }KC_2H_5CO_2\times \text{Molar mass of }KC_2H_5CO_2$

Molar mass of $KC_2H_5CO_2$ = 112 g/mol

$\text{Mass of }KC_2H_5CO_2=0.219mol\times 112g/mol=24.5g$

Thus, the mass of $KC_2H_5CO_2$ is, 24.5 grams