Answer to Question #171488 in Chemistry for Kenan

Question

Calculate the concentration of OH- ions when reacting with 100 cm3 of KOH solution, concentration 0.09 mol / dm3 with 120 cm3 of H2SO4 solution with concentration 0.1 mol / dm3!

Answer 1

The reaction given:

2KOH + H₂SO₄ = K₂SO₄ + 2H₂O

As 1 L = 1dm³ and 1 ml = 1 cm³, the number of moles of the reactants equals:

n(KOH) = M × V = 0.09 mol/L × 100 mL = 0.009 mol

n(H₂SO₄) = M × V = 0.1 mol/L × 120 mL = 0.012 mol

Since 2 mol of KOH are needed to neutralize 1 mole of acid, the number of moles of the acid left in the solution equals:

n(H₂SO4) = 0.012 mol - (0.009 / 2) = 0.0075 mol

The concentration of H+ ions equals:

[H+] = (2 × 0.0075 mol) / (100 ml + 120 ml) = 0.068 M

From here, pH of the solution is:

pH = -log[H+] = 1.17 M

pOH = 14 - 1.17 = 12.83

From here:

[OH-] = 10^-12.83 = 1.5 × 10^-13 M

Answer: 1.5 × 10^-13 M