Question

Calculate the pH after 0.16 mole of NaOH is added to 1.10 L of a solution that is 0.58 M HCO2H and 1.12 M HCO2Li, and calculate the pH after 0.32 mole of HCl is added to 1.10 L of the same solution of HCO2H and HCO2Li.

0.16 mole of NaOH

0.32 mole of HCl

Answer 1

(a): HCOOH and HCOOLi acts as a buffer solution. When NaOH is added it reacts with HCOOH to form HCOO- and H2O.

NaOH + HCOOH ----- > HCOO- + H2O

Moles of HCOOH in the solution = MxV = 0.58 M x 1.10 L = 0.638 mol

Moles of HCOO- in the solution = MxV = 1.12 Mx1.10 L = 1.232 mol

Hence 0.16 mol of NaOH added will react with 0.16 mol of HCOOH to form 0.16 mol of HCOO-.

Moles fo HCOO- after the reaction = 1.232 + 0.16 = 1.392 mol

Hence concentration of HCOO- after the reaction, [HCOO-] = 1.392 mol / 1.10L = 1.2655 M

Moles of HCOOH after the reaction = 0.638 - 0.16 = 0.478 mol

Hence concentration of HCOOH after the reaction, [HCOOH] = 0.478 mol / 1.10L = 0.43455 M

Now applying Hendersen equation

pH = pKa + log[HCOO-] / [HCOOH]

=> pH = 3.77 + log(1.2655 M / 0.43455 M) = 4.234 (answer)

(b) When 0.32 mol of HCl is added it will react with 0.32 mol of HCOO- to form 0.32 mol of HCOOH.

Moles of HCOOH after the reaction = 0.638 + 0.32 = 0.958 mol

Hence concentration of HCOOH after the reaction, [HCOOH] = 0.958 mol / 1.10L = 0.871 M

Moles fo HCOO- after the reaction = 1.232 - 0.32 = 0.912 mol

Hence concentration of HCOO- after the reaction, [HCOO-] = 0.912 mol / 1.10L = 0.8291 M

Now applying Hendersen equation

pH = pKa + log[HCOO-] / [HCOOH]

=> pH = 3.77 + log(0.8291 M / 0.871 M) = 3.749 (answer)