Question

Calculate the pH of a solution made by mixing 25.0 mL of 0.09 M NaOH and 25.0 mL of 0.04 M H₂SO₄.

A) 11.70

B) 11.42

C) 2.30

D) 2.97

E) 10.11

Answer 1

Molarity of NaOH = 0.09M

Volume of NaOH = 25.0 mL

Molarity of H₂SO₄ = 0.04M

Volume of H₂SO₄ = 25.0 mL

From the above equilibrium, it is clear that 1 mol of H₂SO₄ reacts with 2 mol of NaOH.

No. of mol is given by the formula

So, no. of mol of H₂SO₄ is

No. of mol of NaOH is

1 mol of H₂SO₄ reacts with 2 mol of NaOH

Therefore, 1.00 x 10^-3 mol of H₂SO₄ will react with no. of mol of NaOH given by

Therefore, NaOH is in excess and there will be no H₂SO₄ remaining.

Excess NaOH is

1 mol of NaOH gives 1 mol of OH^-

So, no. of mol of OH^- = 0.25 x 10^-3mol

Total volume = 25.0 mL + 25.0 mL = 50.0 mL = 50.0 x 10^-3L

Therefore, concentration of OH^- is

Therefore, pH of the solution is 11.70

Option (a) is the answer.