Question

What is the final pH when 9.77 g of phosphoric acid (98.00 g/mol) and 4.75 g of NaOH (40.00 g/mol) are added to 1 L of water. K_a1 = 1.1 x 10^-2 K_a2 = 7.5 x 10^-8 K_a3= 4.3 x 10^-13

Answer 1

Solution:

Values given in the question is,

Mass of phosphoric acid = 9.77 g; Molar mass of phosphoric acid = 98.00 g/mol

Mass of NaOH = 4.75 g; Molar mass of NaOH = 40.00 g/mol

Volume of water = 1L

K_a1 = 1.1 x 10^-2; K_a2 = 7.5 x 10^-8 and K_a3= 4.3 x 10^-13

Phosphoric acid is a polyprotic acid that can can lose more than one hydrogen. The ionisation occurs stepwise, and each step has its own Ka.

H₃PO₄ (aq) ⇌ H⁺ (aq) + H₂PO₄^- (aq) K_a1 = 1.1 x 10^-2

H₂PO₄^- (aq) ⇌ H⁺ (aq) + HPO₄^2- (aq) K_a2 = 7.5 x 10^-8

HPO₄^2- (aq) ⇌ H⁺ (aq) + PO₄^3- (aq) K_a3= 4.3 x 10^-13

Step 1 Calculating concentration of phosphoric acid & NaOH.

Phosphoric acid:

Number of moles = Mass / Molar mass = 9.77 g / 98 g mol^-1 = 0.0997 mol

Concentration = number of moles / Water volume L = 0.0997 mol / 1L = 0.0997M

NaOH:

Number of moles = Mass / Molar mass = 4.75 g / 40 g mol^-1 = 0.1188 mol

Concentration = number of moles / Water volume L = 0.1188 mol / 1L = 0.1188 M

Step 2 Calculating the concentration of NaOH & H₂PO₄^- after first ionisation stage.

H₃PO₄ (aq) + OH^-(aq) ⇌ H₂PO₄^- (aq) + H₂O

Initial concentration 0.0997 0.1188 0

Change concentration - 0.0997 - 0.0997 0.0997

Equilibrium concentration 0 0.0191 0.0997

Hence after 1st ionisation, concentration of NaOH is 0.0191M and H₂PO₄^- is 0.0997M.

Step 3 Calculating the concentration of NaOH after second ionisation stage.

H₂PO₄^- (aq) + OH^-(aq) ⇌ HPO₄^2- (aq) + H₂O

Initial concentration 0.0997 0.0191 0

Change concentration - 0.0191 - 0.0191 0.0191

Equilibrium concentration 0.0806    0    0.0191

Hence after 2nd ionisation, concentration of NaOH is 0M .

Step 4 Calculating pH.

Since there is no NaOH reaction will not proceed. Further, the Ka3 is negligible we shall calculaate the pH using Ka2.

P^Ka2 = - log Ka2 = - 7.5 x 10^-8 = 7.125

pH = P^Ka2 + log ([HPO₄^2-] / [H₂PO₄^-])

= 7.125 + log (0.0191/0.0806) = 7.125 + log 0.237 = 7.125 + (-0.625) = 7.125 - 0.625 = 6.5

Hence the final pH of the solution is 6.5.