Questions: In which of these five solutions will a precipitate of Ag2CrO4(s) form? Ksp for Ag2CrO4(s)=9.0 x 10^-12. (Select all that apply.) a. 1.0 x 10^-7 M AgNO3 and 6.0 x 10^-4 M K2CrO4 b. 1.0 x 10^-6 M AgNO3 and 6.0 x 10^-4 M K2CrO4 c. 1.0 x 10^-5 M AgNO3 and 6.0 x 10^-4 M K2CrO4 d. 1.0 x 10^-3 M AgNO3 and 6.0 x 10^-4 MK2CrO4 e. 1.0 x 10^-2 M AgNO3 and 6.0 x 10^-4 M K2CrO4

In which of these five solutions will a precipitate of Ag2CrO4(s) form? Ksp for Ag2CrO4(s)=9.0 x 10^-12.
(Select all that apply.)
a. 1.0 x 10^-7 M AgNO3 and 6.0 x 10^-4 M K2CrO4
b. 1.0 x 10^-6 M AgNO3 and 6.0 x 10^-4 M K2CrO4
c. 1.0 x 10^-5 M AgNO3 and 6.0 x 10^-4 M K2CrO4
d. 1.0 x 10^-3 M AgNO3 and 6.0 x 10^-4 MK2CrO4
e. 1.0 x 10^-2 M AgNO3 and 6.0 x 10^-4 M K2CrO4

Transcript text: In which of these five solutions will a precipitate of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}(s)$ form? $K_{\mathrm{sp}}$ for $\mathrm{Ag}_{2} \mathrm{CrO}_{4}(s)=9.0 \times 10^{-12}$. (Select all that apply.) a. $1.0 \times 10^{-7} \mathrm{M} \mathrm{AgNO}_{3}$ and $6.0 \times 10^{-4} \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}$ b. $1.0 \times 10^{-6} \mathrm{M} \mathrm{AgNO}_{3}$ and $6.0 \times 10^{-4} \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}$ c. $1.0 \times 10^{-5} \mathrm{M} \mathrm{AgNO}_{3}$ and $6.0 \times 10^{-4} \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}$ d. $1.0 \times 10^{-3} \mathrm{M} \mathrm{AgNO}_{3}$ and $6.0 \times 10^{-4} \mathrm{MK}_{2} \mathrm{CrO}_{4}$ e. $1.0 \times 10^{-2} \mathrm{M} \mathrm{AgNO}_{3}$ and $6.0 \times 10^{-4} \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}$

Solution

Solution Steps

Step 1: Understand the Problem

We need to determine in which solutions a precipitate of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}(s)$ will form. The solubility product constant ($K_{\mathrm{sp}}$) for $\mathrm{Ag}_{2} \mathrm{CrO}_{4}(s)$ is given as $9.0 \times 10^{-12}$.

Step 2: Write the Expression for $K_{\mathrm{sp}}$

The dissociation of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}(s)$ in water is: \[ \mathrm{Ag}_{2} \mathrm{CrO}_{4}(s) \rightleftharpoons 2\mathrm{Ag}^{+}(aq) + \mathrm{CrO}_{4}^{2-}(aq) \] The expression for the solubility product is: \[ K_{\mathrm{sp}} = [\mathrm{Ag}^{+}]^2 [\mathrm{CrO}_{4}^{2-}] \]

Step 3: Calculate the Ion Product for Each Solution

For each solution, calculate the ion product ($Q$) and compare it to $K_{\mathrm{sp}}$. A precipitate will form if $Q > K_{\mathrm{sp}}$.

Solution a: \[ [\mathrm{Ag}^{+}] = 1.0 \times 10^{-7} \, \mathrm{M}, \quad [\mathrm{CrO}_{4}^{2-}] = 6.0 \times 10^{-4} \, \mathrm{M} \] \[ Q = (1.0 \times 10^{-7})^2 \times (6.0 \times 10^{-4}) = 6.0 \times 10^{-18} \]
Solution b: \[ [\mathrm{Ag}^{+}] = 1.0 \times 10^{-6} \, \mathrm{M}, \quad [\mathrm{CrO}_{4}^{2-}] = 6.0 \times 10^{-4} \, \mathrm{M} \] \[ Q = (1.0 \times 10^{-6})^2 \times (6.0 \times 10^{-4}) = 6.0 \times 10^{-16} \]
Solution c: \[ [\mathrm{Ag}^{+}] = 1.0 \times 10^{-5} \, \mathrm{M}, \quad [\mathrm{CrO}_{4}^{2-}] = 6.0 \times 10^{-4} \, \mathrm{M} \] \[ Q = (1.0 \times 10^{-5})^2 \times (6.0 \times 10^{-4}) = 6.0 \times 10^{-14} \]
Solution d: \[ [\mathrm{Ag}^{+}] = 1.0 \times 10^{-3} \, \mathrm{M}, \quad [\mathrm{CrO}_{4}^{2-}] = 6.0 \times 10^{-4} \, \mathrm{M} \] \[ Q = (1.0 \times 10^{-3})^2 \times (6.0 \times 10^{-4}) = 6.0 \times 10^{-10} \]
Solution e: \[ [\mathrm{Ag}^{+}] = 1.0 \times 10^{-2} \, \mathrm{M}, \quad [\mathrm{CrO}_{4}^{2-}] = 6.0 \times 10^{-4} \, \mathrm{M} \] \[ Q = (1.0 \times 10^{-2})^2 \times (6.0 \times 10^{-4}) = 6.0 \times 10^{-8} \]