Question

10. (a) What is the pH of 0.9mol of acetic acid and 0.8mol sodium hydroxide in 2.0L solution? (b) What will the pH be after 0.010 mol of NaOH has been added to 100.0 mL of this solution?

Answer 1

b.Use the Henderson -Hasslebach equation.

pH = pKa + log [Salt] / [Acid]
So,
pH = pKa + log 1 = pKa

So pKa for HCCOH = 3.77

Now there is a change in the concentration of acid and salt.

In 100 mL, number of moles of acid present = 0.05 M
In 100 mL, number of moles of salt present = 0.05 M

On adding the NaOH, we increase the numeber of moles of Salt and decrease those of the acid

HCOOH + NaOH -----> HCOONa + H2O

So, new number of moles of acid = 0.04
new number of moles of salt = 0.06

Using the equation again, we have
pH = pKa + log [salt]/[acid]
pH = 3.77 + log 0.06/0.04
pH = 3.77 + log 1.5
pH = 3.9460 = 3.95