Question

a solution contains 1.23×10⁻² m calcium acetate and 8.73×10⁻³ m lead nitrate. solid ammonium chromate is added slowly to this mixture. a. what is the formula of the substance that precipitates first? formula = b. what is the concentration of chromate ion when this precipitation first begins? cro₄²⁻ = m

Explanation:

Step1: Identify possible precipitates

Possible precipitates: $\text{CaCrO}_4$ ($K_{sp}=7.1\times10^{-4}$) and $\text{PbCrO}_4$ ($K_{sp}=2.8\times10^{-13}$).

Step2: Calculate $[\text{CrO}_4^{2-}]$ for each

For $\text{CaCrO}_4$: $[\text{CrO}_4^{2-}]=\frac{K_{sp}}{[\text{Ca}^{2+}]}=\frac{7.1\times10^{-4}}{1.23\times10^{-2}}\approx5.77\times10^{-2}\ \text{M}$
For $\text{PbCrO}_4$: $[\text{CrO}_4^{2-}]=\frac{K_{sp}}{[\text{Pb}^{2+}]}=\frac{2.8\times10^{-13}}{8.73\times10^{-3}}\approx3.21\times10^{-11}\ \text{M}$

Step3: Determine first precipitate

$\text{PbCrO}_4$ needs lower $[\text{CrO}_4^{2-}]$, so precipitates first.

Answer:

A. $\text{PbCrO}_4$
B. $3.21\times10^{-11}$