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}=(23g)/(60g/mol)=0.383mol`

and,

`\text{Moles of }CH_3COONa=\frac{\text{Given mass }CH_3COONa}{\text{Molar mass }CH_3COOH}=(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* 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 ([Salt])/([Acid])`

`pH=pK_a+\log ([CH_3COONa])/([CH_3COOH])`

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

`pH=4.7+\log ((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