Question

Calculate the pH of the following solutions. A solution is prepared by mixing equal volumes of 0.15 M HCl and 0.53 M HNO3. (Assume that volumes are additive.)

Answer 1

Let the voume of both the solutions be 1 litres

Now, as both HCl and HNO₃ are both strong acids, so they will dissociate completely into their respective ions

HCl(aq) --------> H⁺(aq) + Cl^-(aq)

HNO₃(aq) ----------> H⁺(aq) + NO₃^-(aq)

Thus, total moles of H⁺ = molarity of HCl*volume of solution in litres + molarity of HNO₃*volume of solution in litres = 0.15*1 + 0.53*1 = 0.68

Thus, new H⁺concentration = moles of H⁺/volume of solution in litres = 0.68/2 = 0.34 M

Thus, pH = -log[H⁺] = 0.469