Q..50ml of 0.001 NHCL and 10ML of 0.01 NH2 SO4 are mixed together calculate PH of resulting solution?

Answer 1

What is the pH of the solution?

50mL 0.001N HCl + 10mL 0.01N H₂SO₄

We will express 50mL as 50e-3L and 10mL as 10e-3L to keep our volume in Liters instead of mL.

Normal Concentration = Molar Concentration x # equivalents/mol

1 equivalent H/mol (HCl)

2 equivalents H/mol (H₂SO₄)

Molar C (HCl) = 0.001N/1 equivalent/mol = 0.001M HCl

0.001mol/L H [50e-3L] = 5e-5 mol H

Molar C (H₂SO₄) = 0.01N/2 equivalents/mol = 0.005M H₂SO₄

0.005 mol/L H₂SO₄ [10e-3L] = 5e-5 mol H₂SO₄

5e-5 mol H (HCl) + (5e-5 mol H x 2) = 0.00015 mol H; This is the total number of moles H⁺ in solution. We multiply the moles of H₂SO₄ by 2 because we have 2 equivalents of H⁺ in one mole of H₂SO₄.

Determine [H⁺] by dividing total number of moles by total volume (in Liters).

0.00015 mol H/(50e-3L + 10e-3L) = 0.00015 mol/60e-3L = 0.0025M H

Input concentration of H+ into pH equation.

pH = -log[H+] = -log[0.0025] = 2.60