Question

How many liters of 0.951 M LiOH will be needed to raise the pH of 0.263 L of 4.55 M carbonic acid (H₂CO₃) to a pH of 9.904?

Answer 1

the dissociation H2CO3 (aq) is as follows:

H2CO3(aq) ------->2H+(aq) + CO3^2-(aq)

calculate the moles of H⁺ from the pH as follows:

[H⁺] = 10^-9.904

= 1.247 x 10^-10

number of moles of H⁺ = 1.247 x 10^-10

calculate the number of moles of H⁺ from initial H2CO3

number of moles of H2CO3 = 4.55 M x 0.263 L

= 1.197 mol

number of moles of H⁺ = 2 x 1.197 = 2.394 mol

number of moles of H⁺ neutralised = 2.394 mol -  1.247 x 10^-10 (very small)

= 2.394 mol

volume of LiOH = 2.394 mol / 0.951 M = 2.517 L