Question

A solution is 0.015 M in both Br– and SO42–. A 0.204 M solution of lead(II) nitrate is slowly added to it with a buret.The ____ anion will precipitate from solution first.(Ksp for PbBr2 = 6.60 ×× 10–6; Ksp for PbSO4 = 2.53 ×× 10–8)What is the concentration in the solution of the first anion when the second one starts to precipitate at 25°C?

Answer 1

The lower Ksp value for PbSO4 indicates that it will precipitate first from a solution containing equal concentrations of Br− and SO4²−. After establishing the solubility equilibrium expressions and calculating the concentrations, we can find the concentration of the first ion when the second precipitate begins to form.

Step-by-step explanation:

To determine which anion will precipitate first when a 0.204 M solution of lead(II) nitrate is slowly added to a 0.015 M solution containing both Br− and SO4²−, we need to use the solubility product constants (Ksp) provided. The anion that will precipitate first is the one that will reach its Ksp value first upon addition of Pb(NO3)2.

The Ksp values are:

• Ksp for PbBr2 = 6.60 × 10−6
• Ksp for PbSO4 = 2.53 × 10−8

The dissolution reactions are:

• PbBr2 (s) → Pb²+ (aq) + 2Br− (aq)
• PbSO4 (s) → Pb²+ (aq) + SO4²− (aq)

To find out which salt precipitates first, we compare the reaction quotient (Q) with the Ksp. The reaction quotient is calculated by multiplying the concentrations of the products. When Q equals Ksp, the solution is saturated and precipitation begins.

For PbBr2:

Q = [Pb²+][Br−]2

For PbSO4:

Q = [Pb²+][SO4²−]

Since the concentrations of Br− and SO4²− are initially equal, PbSO4 will precipitate first because it has a much lower Ksp value. Therefore, sulfate ions will precipitate as PbSO4 first, and the concentration of sulfate will be very low when lead bromide starts to precipitate.

The concentration of Br− when PbSO4 starts to precipitate can be found by setting up the equilibrium expression for PbSO4 and solving for [Pb²+], then substituting [Pb²+] into the Ksp expression for PbBr2 and solving for [Br−].

Answer 2

The first anion to precipitate when lead(II) nitrate is added to a solution containing Br⁻ and SO4²⁻ is SO4²⁻, due to its lower Ksp value compared to PbBr2. The concentration of the SO4²⁻ anion remains constant at 0.015 M once PbSO4 begins to precipitate.

Step-by-step explanation:

The precipitation of anions when a solution of lead(II) nitrate is added depends on the solubility products (Ksp) of the potential precipitates.

The Ksp for PbBr2 is 6.60 × 10⁻⁶, and the Ksp for PbSO4 is 2.53 × 10⁻⁸. To find the anion that precipitates first, we compare the ion product ([Pb²⁺][anion⁻²]) with the Ksp values. Since the Ksp of PbSO4 is smaller, it will precipitate first when the product of [Pb²⁺] and [SO4²⁺] exceeds 2.53 × 10⁻⁸. After the precipitation of PbSO4 starts, no further change in Br⁻ concentration occurs.

To find the concentration of the first anion, which is SO4²-, when PbBr2 starts to precipitate, we must first reach the point at which Ksp of PbSO4 is equaled, and then determine the concentration of lead ions at that point. Because PbBr2 has a higher Ksp, it will start to precipitate at a higher concentration of lead ions. At the moment of first precipitation of PbSO4, the concentration of SO4²- will remain constant at 0.015 M.