Question

A 56.5 ml sample of a 0.122 M potassium sulfate solution is mixed with 35.0 ml of a 0.128 M lead (II) acetate solution, and the following precipitation reaction occurs.

a. K2SO4 + Pb(CH3COO)2 → 2 K(CH3COO) + PbSO4
b. K2SO4 + Pb(CH3COO)2 → 2 K(CH3COO)2 + PbSO4
c. K2SO4 + Pb(CH3COO)2 → K2(CH3COO)2 + PbSO4
d. 2 K2SO4 + Pb(CH3COO)2 → 2 K(CH3COO) + PbSO4

Answer 1

The correct precipitation reaction between potassium sulfate and lead (II) acetate is represented by option (a), which results in soluble potassium acetate and insoluble lead (II) sulfate.

Step-by-step explanation:

To identify which precipitation reaction occurs when potassium sulfate is mixed with lead (II) acetate, let's consider the solubility rules. Sulfates are generally soluble except when paired with lead, barium, or strontium ions. Therefore, a precipitate of lead (II) sulfate will form. Acetate salts are generally soluble, so the potassium acetate will remain in solution. This leads us to the correct balanced equation:

K2SO4(aq) + Pb(CH3COO)2(aq) → 2 KCH3COO(aq) + PbSO4(s)

This matches option (a), which is the balanced reaction. The other options are either not balanced properly or predict products that violate solubility rules, as potassium does not form precipitates with the acetate ion.