Question

Calculate the pH of a solution made by adding 59 g of sodium acetate, NaCH3COO, to 23 g of acetic acid, CH3COOH, and dissolving in water to make 400. mL of solution. Hint given in feedback.

Answer 1

This is an incomplete question, here is a complete question.

Calculate the pH of a solution made by adding 59 g of sodium acetate, NaCH₃COO, to 23 g of acetic acid, CH₃COOH, and dissolving in water to make 400. mL of solution. Hint given in feedback. The Ka for CH₃COOH is 1.8 x 10⁻⁵ M. As usual, report pH to 2 decimal places.

Answer : The pH of the solution is, 4.97

Explanation :

First we have to calculate the moles of CH₃COOH and NaCH₃COO

$\text{Moles of }CH_3COOH=\frac{\text{Given mass }CH_3COOH}{\text{Molar mass }CH_3COOH}=\frac{23g}{60g/mol}=0.383mol$

and,

$\text{Moles of }CH_3COONa=\frac{\text{Given mass }CH_3COONa}{\text{Molar mass }CH_3COOH}=\frac{59g}{82g/mol}=0.719mol$

Now 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:

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

$pK_a=5-\log (1.8)$

$pK_a=4.7$

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

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

$pH=pK_a+\log \frac{[CH_3COONa]}{[CH_3COOH]}$

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

$pH=4.7+\log (\frac{0.719}{0.383})$

(As the volume is same. So, we can write concentration in terms of moles.)

$pH=4.97$

Therefore, the pH of the solution is, 4.97