Calculate the resulting pH of each of the following mixtures.

Question

Calculate the resulting pH of each of the following mixtures.

a. 100. mL of 0.100 M Na₃PO₄ + 150. mL of 0.180 M HNO₃

b. 100 mL of 0.200 M NH₃ + 50.0 mL of 0.500 M HCl

Answer 1

Na₃PO₄ + 3HNO₃ ==> H₃PO₄ + 3NaNO₃

moles Na₃PO₄ = 100. ml x 1 L / 1000 ml x 0.100 mol / L = 0.0100 mols

moles HNO3 = 150. ml x 1 L / 1000 ml x 0.180 mol / L = 0.027 mols

Na₃PO₄ + 3HNO₃ ==> H₃PO₄ + 3NaNO₃

0.01........0.027...............0...............0.............Initial

-0.009....-0.027..............+0.009.....+0.027.....Change

0.001.........0..................0.009..........0.027....Equilibrium

Final volume = 100 ml + 150 ml = 250 ml = 0.250 L

Final concentrations:

[Na3PO4] = 0.001 mol / 0.250 L = 4x10^-3 M

[H3PO4] = 0.009 mol / 0.250 L = 0.036 M

[NaNO3] = 0.027 mol / 0.250 L = 0.108 M

Na3PO4 wil not affect pH very much upon hydrolysis

NaNO3 will be neutral upon hydrolysis

pH will mostly be determined by H3PO4 and we will only consider the first H of H3PO4:

Ka = 7.5x10^-3 = [H+][H2PO4] / [H3PO4] = (x)(x)/0.036-x

x² = 2.7x10^-4 - 7.5x10^-3x

Solve for x, take negative log to get pH

NH3 + HCl ==> NH4Cl

mols NH3 = 100 ml x 1 L / 1000 ml x 0.200 mol/L = 0.0200 mols NH3

mols HCl = 50.0 ml x 1 L / 1000 ml x 0.500 mol/L = 0.025 mols HCl

excess HCl = 0.025 mol - 0.0200 mol = 0.00500 mols HCl

Final volume = 150 ml = 0.150 L

Final [HCl] = 0.00500 mol / 0.150 L = 0.0333 M

pH = -log [H+] = -log 0.0333

pH = 1.48