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

and,

Now we have to calculate the value of
.
The expression used for the calculation of
is,

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



Now we have to calculate the pH of the solution.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[CH_3COONa]}{[CH_3COOH]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D)
Now put all the given values in this expression, we get:

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

Therefore, the pH of the solution is, 4.97