Which of the following is the correct equation for the K_sp for Fe₃(PO₄)₂ ?

Solution:

K_sp = [Fe²⁺]³ [PO₄^3-]²

What is the molar solubility of Fe(OH)₂ in a solution buffered at a pH of 9.40 ?
The K_sp for iron(II) hydroxide is 4.9 × 10^-17 .

Solution:

The fact that the solution is buffered at a pH of 9.40 (pOH = 4.60) means that the concentration of hydroxide (OH^-) is set. You only have to solve for the iron(II) ion concentration and that will be equal to the molar solubility.

Convert pH to concentration of hydroxide: [OH^-] = 10^-pOH = 10^-4.6 = 2.5 × 10^-5 M

Rearrange the solubility product equation:
[Fe²⁺][OH^-]² = K_sp
[Fe²⁺] = K_sp/[OH^-]²
[Fe²⁺] = 4.9 × 10^-17/(2.5 × 10^-5)²
[Fe²⁺] = 7.8 × 10^-8 M

What will be the pH at the equivalence point of a titration of 0.30 M hydrofluoric acid (HF) with an equimolar solution of NaOH? hydrofluoric acid has a K_a of 6.3 × 10⁻⁴.

Solution:

The reaction is:

HF + NaOH → H₂O + NaF

Note that because equimolar concentrations were used, the volume of base (NaOH) used to neutralize the HF will equal the volume of the HF. This means that the original concentration will be halved due to the doubling of the volume. Therefore, the concentration of the conjugate base (F^-) will be 0.15 M at the equivalence point. Now work like any weak base solution.

Convert K_a into K_b
1.0 × 10^-14 / 6.3 × 10^-4 = 1.59 × 10^-11

Use the assumption and get hydroxide concentration:
[OH^-] = \(\sqrt{K_{\rm b}\cdot C_{\rm F-}}\) = 1.54 × 10^-6 M

Now convert OH^- concentration into pOH
pOH = -log(1.54 × 10^-6) = 5.812

Now convert pOH into pH
pH = 14 - pOH = 14 - 5.812 = 8.188 = 8.19

40.0 mL of an HBr solution with a pH of 2.77 neutralizes 260 mL of a Ba(OH)₂ solution. What is the molarity of the Ba(OH)₂ solution?

Solution:

The reaction is:

2 HBr + Ba(OH)₂ → 2 H₂O + BaBr₂

Convert pH to concentration: [H⁺] = 10^-pH = 10^-2.77 = 1.7 × 10^-3 M

Now convert to concentration to moles of HBr
1.7 × 10^-3 mol/L × 40 mL = 0.068 mmol HBr

Convert mmol of HBr to mmol of Ba(OH)₂
0.068 mmol (1 mmol Ba(OH)₂ / 2 mmol HBr) = 0.034 mmol Ba(OH)₂

Convert mmol Ba(OH)₂ to concentration
0.034 mmol Ba(OH)₂/ 260 mL = 1.3 × 10^-4 M