0.05g of magnesium react in 23.92mL of 2M formic acid (HCOOH, Ka=1.8x10-4). Determine the resulting pH.

Question

0.05g of magnesium react in 23.92mL of 2M formic acid (HCOOH, K_a=1.8x10^-4). Determine the resulting pH.

Answer 1

Write the reaction taking place:

Mg(s) + 2HCOOH(aq) ==> Mg²⁺(aq) + 2HCOO^- + H2(g)

Next, find the limiting reactant:

0.05 g Mg x 1 mol Mg/24 g x 2 mol HCOO- / mol Mg = 0.0042 mol HCOO-

23.92 ml formic acid x 1 L/1000 ml x 2 mol/L x 2 mol HCOO- / 2 mol HCOOH = 0.048 mol HCOO-

Therefore Mg is limiting and the amount of HCOO- formed will be 0.0042 moles. This will be in 23.92 ml.

Now we can calculate the concentration of HCOO- and then look at the hydrolysis of this ion as follows:

[HCOO-] = 0.0048 mol/0.02392 L = 0.200 M

Hydrolysis: HCOO- + H2O ==> HCOOH + OH-

HCOO- is acting as a base, so we need the Kb for HCOO-. This we get from the given Ka for HCOOH...

KaKb = 1x10^-14 and Kb = 1x10^-14/Ka = 1x10^-14 / 1.8x10^-4 and Kb = 5.6x10^-11

Kb = 5.6x10^-11 = [HCOOH[OH-] / [HCOO-] = (x)(x) / 0.200

x² = 1.1x10^-11

x = 1.06x10^-6 M = [OH-]

pOH = -log 1.06x10^-6 = 5.98

pH = 14 - pOH

pH = 8.02