Question

what is the pH of a solution following the combination of 245.0 mL of a 4.00*10^-3 M NaOH solution with 255.0 mL of a 2.33*10^-3 M H2SO4 solution?

Answer 1

no. of mole = molarity volume of solution in liter

mole of NaOH = (4.0010^-3)(0.245L) = 0.00098 mole

mole of H₂SO₄ = (2.3310^-3)(0.255L) = 0.00059415 mole

reaction of NaOH with H₂SO₄ take place as follow

2NaOH + H₂SO₄ Na₂SO^2-₄ + 2H₂O

According to reaction 2 mole of NaOH react with 1 mole of H₂SO₄

then 0.00098 mole of NaOH react with 0.00098 / 2 = 0.00049 mole of H₂SO₄

mole of H₂SO₄ remained unreacted = 0.00059415 - 0.00049 = 0.00010415 mole

H₂SO₄ dissociated as

H₂SO₄ 2H⁺ +SO^2-₄

1 mole H₂SO₄ produce 2 mole H⁺ ion then 0.00010415 mole of H₂SO₄ produce 0.000104152 = 0.00020830 mole of H⁺ ion

total volume of solution = 245ml + 255ml = 500 ml = 0.5L

malarity = no. of mole / volume of solution in liter

[H⁺] = 0.00020830/0.5 = 0.0004166

pH = -log[H⁺] = -log(0.0004166) = 3.38

pH of solution = 3.38