Question

What amount of solid NaOH must be added to 1.0 L of a 0.15 M H2CO3 solution to produce a solution with [H+]= 3.0×10⁻¹¹ M?

There is no significant volume change as the result of the addition of the solid.

Answer 1

[H⁺] = 3.0 x 10^-11 M

[OH^-] = K_w / [H⁺]

[OH^-] = (1.0 x 10^-14) / (3.0 x 10^-11 M)

[OH^-] = 3.3 x 10^-4 M

moles OH^- in solution = (molarity OH^-) * (volume of solution in Liter)

moles OH^- in solution = (3.3 x 10^-4 M) * (1.0 L)

moles OH^- in solution = 3.3 x 10^-4 mol

moles H₂CO₃ = (molarity H₂CO₃) * (volume of solution in Liter)

moles H₂CO₃ = (0.15 M) * (1.00 L)

moles H₂CO₃ = 0.15 mol

Reaction between H₂CO₃ and NaOH is

H₂CO₃ + 2 NaOH Na₂CO₃ + 2 H₂O

moles NaOH reacted with H₂CO₃ = 2 * (moles H₂CO₃)

moles NaOH reacted with H₂CO₃ = 2 * (0.15 mol)

moles NaOH reacted with H₂CO₃ = 0.30 mol

Total moles NaOH added = (moles OH^- in solution) + (moles NaOH reacted with H₂CO₃)

Total moles NaOH added = (3.3 x 10^-4 mol) + (0.30 mol)

Total moles NaOH added = 0.30033 mol

mass NaOH added = (Total moles NaOH added) * (molar mass NaOH)

mass NaOH added = (0.30033 mol) * (40 g/mol)

mass NaOH added = 12 g