4.5 mL of 0.0750 M KOH is added to a 32.0 mL solution of 0.025 M...

Question

Question

4.5 mL of 0.0750 M KOH is added to a 32.0 mL solution of 0.025 M HNO2. Calculate the pH of the solution.

Answer 1

Answer #1

KOH + HNO₂ ------> KNO₂ + H₂O

moles of KOH = molrity*volume of solution in litres = 0.075*0.0045 = 3.375*10^-4

moles of HNO₂ = molrity*volume of solution in litres = 0.025*0.032 = 8*10^-4

Thus, excess HNO₂ = (8*10^-4) - (3.375*10^-4) = 4.625*10^-4

Now,

HNO₂(aq) -------> H⁺(aq) + NO₂^-(aq)

[H⁺] = 4.625*10^-4 M

pH = -log[H⁺] = 3.335

Answer 2

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