Question

What is the pH of a solution made by mixing 100.0 mL of 0.10 M HNO3, 50.0 mL of 0.20 M HCl, and 100.0 mL of water? Assume that the volumes are additive.

must show work

Answer 1

no. of mole = molarity X volume of solution in liter

no.of mole of HNO₃ = 0.10 X 0.100 L = 0.01 mole

no. of mole of HCl = 0.2 X 0.05 L = 0.01 mole

both HNO₃ and HCl are strong acid dissociate completly as given below

HNO₃ + H₂O H₃O⁺ + NO₃^-

HCl + H₂O H₃O⁺ + Cl^-

therefore no. of mole of H₃O⁺ formed by HNO₃ = 0.01 mole

no. of mole of H₃O⁺ formed by HCl = 0.01 mole

total mole of H₃O⁺ = 0.01 + 0.01 = 0.02 mole

total volume of soltuion = 100 + 50 + 100 = 250 ml = 0.250 liter

Molarity = no. of mole / volume of solution in liter

molarity of H₃O⁺ = 0.02 / 0.250 = 0.08 M

pH = -log[H₃O⁺] = -log(0.08) = 1.097

pH of solution = 1.097