Question

1). Both H2PO3− and HS− are amphoteric. Write an equation to show how H2PO3− can act as a base with HS− acting as an acid. Express your answer as a chemical equation. Identify all of the phases in your answer.

2). Determine the pH of each of the following solutions.

a). 8.59×10⁻² M HClO4 b). a solution that is 4.7×10⁻² M in HClO4 and 5.1×10⁻² M in HCl c). a solution that is 1.10% HCl by mass (Assume a density of 1.01 g/mL for the solution.)

3). Determine the pH of an HF solution of each of the following concentrations.

a). 0.270 M b). 5.10×10⁻² M c). 2.30×10⁻² M d). In which cases can you not make the simplifying assumption that x is small?

4). What volume of 0.879 M KOH solution is required to make 3.63 L of a solution with pH of 12.8? Express your answer with the appropriate units.

Answer 1

1) pH = 12.8

Thus, pOH = 14 - pH = 1.2

Now, [OH^-] = 10^-pOH = 0.063

Now, KOH ---> K⁺(aq) + OH^-(aq)

Thus, [KOH] = 0.063 M

Now, since the moles of KOH remains constant after dilution , therefore applying M₁*V₁ = M₂*V₂

or, 0.879*V₁ = 0.063*3.63

or, V₁ = volume required = 0.229 litres

3) K_a of HF = 6.3*10^-4

Now,

HF(aq) <-----> H⁺(aq) + F^-(aq)

a) Initially, [HF] = 0.27 M ; [H⁺] = [F^-] = 0 M

Let at eqb., [HF] = 0.27-x M ; [H⁺] = [F^-] = x M

Thus, K_a = {[H⁺]*[F^-]}/[HF]

or, 6.3*10^-4 = x²/(0.27-x)

or, x = 0.0127 = [H⁺]

Thus, pH = -log[H⁺] = 1.896

b) Initially, [HF] = 0.051M ; [H⁺] = [F^-] = 0 M

Let at eqb., [HF] = 0.051-x M ; [H⁺] = [F^-] = x M

Thus, K_a = {[H⁺]*[F^-]}/[HF]

or, 6.3*10^-4 = x²/(0.051-x)

or, x = 0.00536 = [H⁺]

Thus, pH = -log[H⁺] = 2.271

c) Initially, [HF] = 0.023 M ; [H⁺] = [F^-] = 0 M

Let at eqb., [HF] = 0.023-x M ; [H⁺] = [F^-] = x M

Thus, K_a = {[H⁺]*[F^-]}/[HF]

or, 6.3*10^-4 = x²/(0.023-x)

or, x = 0.0035 = [H⁺]

Thus, pH = -log[H⁺] = 2.456

1) H₂PO₃^-(aq) + HS^-(aq) <----> H₃PO₃(aq) + S^2-(aq)

2) a) HClO₄(aq) -----> H⁺(aq) + ClO₄^-(aq)

Thus, [H⁺] = [HClO₄] = 0.859

pH = -log[H⁺] = 0.066

b) HClO₄(aq) -----> H⁺(aq) + ClO₄^-(aq)

Thus, [H⁺] = [HClO₄] = 0.047 .....(1)

Also, HCl(aq) -----> H⁺(aq) + Cl^-(aq)

Thus, [H⁺] = [HCl] = 0.051.....(2)

Total [H⁺] = 0.098

Thus, pH = -log[H⁺] = 1.009