Question

Enter your answer in the provided box. Calculate the pH of 1.00 L of a buffer that is 1.00 M in acetic acid and 1.00 M in sodium acetate after the addition of 0.450 mole of NaOH.

Answer 1

Answer : The pH of buffer is, 5.17

Explanation : Given,

`pK_a=4.75`

Concentration of acetic acid = 1.00 M

Concentration of sodium acetate = 1.00 M

Volume of solution = 1.00 L

As,
`Moles=Concentration* Volume`

So,

Moles of acetic acid = 1.00 mol

Moles of sodium acetate = 1.00 mol

Moles of NaOH added = 0.450 mol

The balanced chemical equilibrium reaction is:

`CH_3COO+NaOH\rightleftharpoons CH_3COONa+H_2O`

Initial mole 1 0.450 1

At eqm. (1-0.450) 0 (1+0.450)

= 0.55 =1.450

Now we have to calculate the pH of buffer.

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=4.75+\log ((1.450)/(0.55))`

`pH=5.17`

Therefore, the pH of buffer is, 5.17