Question

Calculate the pH of a solution prepared by mixing

a) 25.0 mL of 0.512 M NaOH and 34.0 mL of 0.187 M HCl

b) 46.0 mL of 0.235 M KOH and 50.0 mL of 0.420 M HC₃H₅O₂

c) 400 mL of 0.250 M NH₃ and 250 mL of 0.120 M HCl

Answer 1

a) NaOH(aq) + HCl(aq) ---------> NaCl(s) + H₂O(l)

As per the balnced reaction NaOH and HCl reacts in the molar ratio of 1:1

Thus, moles of HCl present initially = molarity*volume of solution in litres = 0.187*0.034 = 0.00636

moles of NaOH present initially = molarity*volume of solution in litres = 0.512*0.025 = 0.0128

Clearly, NaOH is in excess of 0.00644 moles

Thus, NaOH(aq) ------> Na⁺(aq) + OH^-(aq)

[OH^-] = 0.0064/0.059 = 0.108 M

pOH = -log[OH^-] = 0.965

Thus, pH = 14 - pOH = 13.035

b) C₂H₅COOH(aq) + KOH(aq) -------> C₂H₅COOK(aq) + H₂O(l)

As per the balanced reaction C₂H₅COOH and KOH reacts in the molar ratio of 1:1

Moles of C₂H₅COOH present initially = molarity*volume of solution in litres = 0.42*0.05 = 0.021

Moles of KOH present initially = molarity*volume lof solution in litres = 0.235*0.046 = 0.0108

Clearly, C₂H₅COOH is in excess of 0.0102 moles

Thus, [H⁺] = 0.0102/0.096 = 0.1063 M

pH = -log[H⁺] = 0.974

c) HCl(aq) + NH₃(aq) ------> NH₄Cl(s) + H₂O(l)

A per the balanced reaction HCl & NH₃ reacts in the molar ratio of 1:1

Thus, moles of HCl present initially = molarity*volume of solution in litres = 0.12*0.25 = 0.03

Moles of NH₃ present initially = molarity*volume of solution in litres = 0.1

Clearly, NH₃ is in excess of 0.097 moles

Now, NH₃(aq) -----> NH₄⁺(aq) + OH^-(aq)

Now, [OH^-] = 0.097/0.65 = 0.149 M

Thus, pOH = -log[OH^-] = 0.826

Thus, pH = -log[H⁺] = 13.174