Question

When an excess of PbI2 is placed in 1.0 L of 0.05 M KNO3 solution:

Calculate the pH of the titration solution after addition of 49.00 mL 0.1000 M HCl to 100.0 mL 0.0500 M solution of NaOH:

Answer 1

Concentration HCl = 0.1000 M

volume HCl = 49.0 mL

moles HCl added = (Concentration HCl) * (volume HCl)

moles HCl added = (0.1000 M) * (49.0 mL)

moles HCl added = 4.90 mmol

moles H⁺ = moles HCl added

moles H⁺ = 4.90 mmol

concentration NaOH = 0.0500 M

volume NaOH = 100.0 mL

moles NaOH added = (concentration NaOH) * (volume NaOH)

moles NaOH added = (0.0500 M) * (100.0 mL)

moles NaOH added = 5.00 mmol

moles OH^- = moles NaOH added

moles OH^- = 5.00 mmol

Since moles OH^- > moles H⁺, therefore, solution will be basic

excess moles OH^- = (moles OH^-) - (moles H⁺)

excess moles OH^- = (5.00 mmol) - (4.90 mmol)

excess moles OH^- = 0.100 mmol

total volume = (volume HCl) + (volume NaOH)

total volume = (49.00 mL) + (100.0 mL)

total volume = 149.0 mL

[OH^-] = (excess moles OH^-) / (total volume)

[OH^-] = (0.100 mmol) / (149.0 mmol)

[OH^-] = 6.71 x 10^-4 M

pOH = -log[OH^-]

pOH = -log(6.71 x 10^-4 M)

pOH = 3.17

pH = 14 - pOH

pH = 14 - 3.17

pH = 10.83