pH of a mixture of solutions

Question

Q1. what will be the pH if I add 150ml of 0.1NHCl to 50ml 0.2M Na₃PO₄

Q2. what will be the pH if I add 150ml of 0.1NHCl to 50ml 50mM Na₃PO₄

Answer 1

PO₄^3- + 3H⁺ ==> H₃PO₄ (a weak acid with a Ka1 = 7.1x10-3)

Which reactant is limiting?

HCl: 0.150 L x 0.1 mol/L = 0.0150 mol HCl x 1 mol H₃PO₄/3 mol HCl = 0.005 moles H₃PO₄ formed

PO₄^3-: 0.050 L x 0.2 mol/L = 0.01 mol PO₄^3- x 1 mol H3PO4/mol PO₄^3- = 0.01 mol H₃PO₄ formed

HCl is limiting so you have 0.005 moles H₃PO₄ being formed

Final volume = 200 ml = 0.2 L

Final [H₃PO₄] = 0.005 moles/0.2 L = 0.025 M

H₃PO₄ <==> H⁺ + H₂PO₄ Ka1 = 7.1x10^-3

7.1x10^-3 = (x)(x)/0.025-x

x² + 7.1x10^-3x - 1.775x10^-4 = 0

x = 0.010 M = [H⁺]

pH = -log [H⁺]

pH = 2

For the second part, repeat using same for HCl, but PO₄^3- will be...

0.05 L x 0.05 mol/L = 0.025 moles PO₄^3-

HCl: 0.150 L x 0.1 mol/L = 0.0150 mol HCl x 1 mol H₃PO₄/3 mol HCl = 0.005 moles H₃PO₄ formed

PO₄^3-: 0.05 L x 0.025 mol/L = 0.00125 mole PO₄^3- = 0.00125 moles H₃PO₄ formed

In this case PO₄^3- is limiting and you have only 0.00125 moles H₃PO₄ being formed

This leaves excess HCl which will contribute largely to the pH and we can ignore the contribution of H₃PO₄

0.00125 moles PO₄^3- x 3 moles H⁺/mole H₃PO₄ = 0.00375 moles H⁺ used

moles H⁺ remaining = 0.005 - 0.00375 = 0.00125 moles H⁺

Final volume = 200 ml = 0.2 L

Final [H⁺] = 0.00125 mol/0.2 L = 0.00625 M H⁺

pH = -log [H+]

pH = 2.20