Question

Calculate the change in pH when 7.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).

Calculate the change in pH when 7.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Answer 1

Original moles of NH4+ and NH3 = 0.100 M X 0.1L = 0.0100 mol
Moles HCl added = 07 X 0.100 mol/L = 7X10^-4 mol

So, after HCl addition, moles of NH4+ increases up to 0.0107, and moles of NH3 goes down to 0.0093

pKa NH4+ - 9.25, so the original solution will have had a pH = 9.25

Adding exactly the same amount of base will raise the pH by the same amount, +0.06 pH units, so the final pH will be 9.31

After the addition,
pH = pKa + log (0.0093/0.0107)
pH = 9.19

Therefore, the change in pH = -0.06