Question

practice test f2 (pg 2 of 6) solubility equilibrium, $k_{sp}$ 9. on the basis of the $k_{sp}$ values listed below, what is the order of least soluble to most soluble for these compounds? $\ce{agbr}, k_{sp}=5.4\times10^{-13}$ $\ce{ag2co3}, k_{sp}=8.0\times10^{-12}$ $\ce{agcl}, k_{sp}=1.8\times10^{-10}$ a. $\ce{agbr} < \ce{ag2co3} < \ce{agcl}$ b. $\ce{agbr} < \ce{agcl} < \ce{ag2co3}$ c. $\ce{agcl} < \ce{ag2co3} < \ce{agbr}$ d. $\ce{ag2co3} < \ce{agbr} < \ce{agcl}$

Explanation:

Step1: Recall \( K_{sp} \) and solubility

For sparingly soluble salts, a smaller \( K_{sp} \) generally means lower solubility (but we must consider the stoichiometry for salts with different ion ratios). However, for these three, we can first compare the \( K_{sp} \) magnitudes directly (after checking ion ratios, but here AgBr and AgCl have 1:1 ratio, \( Ag_2CO_3 \) has 2:1 ratio. Wait, actually, let's calculate molar solubility (\( s \)) for each to be precise.

For \( AgBr \) (1:1 salt, \( AgBr(s) ightleftharpoons Ag^+(aq) + Br^-(aq) \)):
\( K_{sp} = [Ag^+][Br^-] = s \cdot s = s^2 \)
\( s = \sqrt{K_{sp}} = \sqrt{5.4 \times 10^{-13}} \approx \sqrt{54 \times 10^{-14}} \approx 7.35 \times 10^{-7} \, M \)

For \( Ag_2CO_3 \) (2:1 salt, \( Ag_2CO_3(s) ightleftharpoons 2Ag^+(aq) + CO_3^{2-}(aq) \)):
\( K_{sp} = [Ag^+]^2[CO_3^{2-}] = (2s)^2 \cdot s = 4s^3 \)
\( s = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{8.0 \times 10^{-12}}{4}} = \sqrt[3]{2.0 \times 10^{-12}} \approx 1.26 \times 10^{-4} \, M \)

For \( AgCl \) (1:1 salt, \( AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq) \)):
\( K_{sp} = [Ag^+][Cl^-] = s \cdot s = s^2 \)
\( s = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} \, M \)

Step2: Compare molar solubilities

Now, compare the \( s \) values:
- \( AgBr \): \( \approx 7.35 \times 10^{-7} \, M \)
- \( Ag_2CO_3 \): \( \approx 1.26 \times 10^{-4} \, M \)
- \( AgCl \): \( \approx 1.34 \times 10^{-5} \, M \)

Order of least soluble (smallest \( s \)) to most soluble (largest \( s \)):
\( AgBr \) (smallest \( s \)) < \( AgCl \) (next) < \( Ag_2CO_3 \)? Wait, no, wait my calculation for \( Ag_2CO_3 \) was wrong? Wait, no: \( 8.0 \times 10^{-12} / 4 = 2.0 \times 10^{-12} \), cube root of \( 2.0 \times 10^{-12} \) is \( \approx 1.26 \times 10^{-4} \), which is larger than \( AgCl \)'s \( 1.34 \times 10^{-5} \). Wait, but initial \( K_{sp} \) values: \( AgBr \) is \( 5.4 \times 10^{-13} \), \( Ag_2CO_3 \) is \( 8.0 \times 10^{-12} \), \( AgCl \) is \( 1.8 \times 10^{-10} \). Wait, maybe I messed up the stoichiometry approach. Alternatively, for 1:1 salts (AgBr, AgCl), smaller \( K_{sp} \) means smaller solubility. So \( AgBr \) (\( 5.4 \times 10^{-13} \)) has smaller \( K_{sp} \) than \( AgCl \) (\( 1.8 \times 10^{-10} \)), so \( AgBr \) is less soluble than \( AgCl \). Now for \( Ag_2CO_3 \), even though its \( K_{sp} \) is \( 8.0 \times 10^{-12} \) (larger than \( AgBr \)'s, smaller than \( AgCl \)'s), its molar solubility is higher because of the 2:1 ratio. Wait, let's recalculate \( Ag_2CO_3 \) solubility:

\( K_{sp} = 4s^3 = 8.0 \times 10^{-12} \)
\( s^3 = 2.0 \times 10^{-12} \)
\( s = \sqrt[3]{2.0 \times 10^{-12}} \approx 1.26 \times 10^{-4} \, M \)

For \( AgCl \): \( s = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} \, M \)

Ah, so \( Ag_2CO_3 \) has higher solubility than \( AgCl \)? Wait, that can't be. Wait, no, \( 1.26 \times 10^{-4} \) is 0.000126, and \( 1.34 \times 10^{-5} \) is 0.0000134. So yes, \( Ag_2CO_3 \) is more soluble than \( AgCl \). Wait, but the \( K_{sp} \) of \( Ag_2CO_3 \) is \( 8.0 \times 10^{-12} \), which is smaller than \( AgCl \)'s \( 1.8 \times 10^{-10} \), but because of the 2:1 ion ratio, its solubility is higher. So the order of least soluble to most soluble:

Least soluble: \( AgBr \) (smallest \( K_{sp} \) among 1:1, and even with 2:1, its solubility is lower than \( Ag_2CO_3 \)) → then \( Ag_2CO_3 \)? Wait no, wait my calculation for \( Ag_2CO_3 \) is wrong? Wait, no, let's check again. Wait, \( Ag_2CO_3 \): \( K_{sp} = [Ag^+]^2[CO_3^{2-}] \). Let \( s \) be solubility, then \( [Ag^+] = 2s \), \( [CO_3^{2-}] = s \). So \( K_{sp} = (2s)^2 \times s = 4s^3 \). So \( s = (K_{sp}/4)^{1/3} \). For \( K_{sp} = 8.0 \times 10^{-12} \), \( s = (8.0 \times 10^{-12}/4)^{1/3} = (2.0 \times 10^{-12})^{1/3} \). \( 2.0 \times 10^{-12} = 20 \times 10^{-13} \)? No, \( 2.0 \times 10^{-12} = 2 \times 10^{-12} \). The cube root of \( 2 \times 10^{-12} \) is \( \approx 1.26 \times 10^{-4} \), which is 0.000126 M. For \( AgCl \), \( s = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} \) M (0.0000134 M). So \( Ag_2CO_3 \) is more soluble than \( AgCl \). Then \( AgBr \) has \( s = \sqrt{5.4 \times 10^{-13}} \approx 7.35 \times 10^{-7} \) M (0.000000735 M), which is less than both. So the order from least to most soluble is \( AgBr < AgCl < Ag_2CO_3 \)? Wait no, wait \( Ag_2CO_3 \) is more soluble than \( AgCl \), so the order is \( AgBr \) (least) < \( AgCl \) < \( Ag_2CO_3 \) (most)? But wait the options:

a. \( AgBr < Ag_2CO_3 < AgCl \)
b. \( AgBr < AgCl < Ag_2CO_3 \)
c. \( AgCl < Ag_2CO_3 < AgBr \)
d. \( Ag_2CO_3 < AgBr < AgCl \)

Wait, my calculation for \( Ag_2CO_3 \) solubility is higher than \( AgCl \), so \( Ag_2CO_3 \) is more soluble than \( AgCl \). So the order from least to most is \( AgBr \) (least) < \( AgCl \) < \( Ag_2CO_3 \) (most), which is option b? Wait no, wait I think I made a mistake in the stoichiometry. Wait, no: \( AgCl \) is 1:1, \( K_{sp} = 1.8 \times 10^{-10} \), so \( s = \sqrt{1.8e-10} \approx 1.34e-5 \) M. \( Ag_2CO_3 \): \( K_{sp} = 8.0e-12 \), \( s = (8.0e-12 / 4)^{1/3} = (2.0e-12)^{1/3} \approx 1.26e-4 \) M. So \( 1.26e-4 \) is larger than \( 1.34e-5 \) (since \( 1.26e-4 = 12.6e-5 \), which is larger than \( 1.34e-5 \)). So \( Ag_2CO_3 \) is more soluble than \( AgCl \). Then \( AgBr \) has \( s = \sqrt{5.4e-13} \approx 7.35e-7 \) M, which is less than both. So the order is \( AgBr < AgCl < Ag_2CO_3 \), which is option b.

Wait, but let's check the \( K_{sp} \) values again. \( AgBr \): \( 5.4e-13 \), \( Ag_2CO_3 \): \( 8.0e-12 \), \( AgCl \): \( 1.8e-10 \). So \( 5.4e-13 < 8.0e-12 < 1.8e-10 \). But for 1:1 salts, \( K_{sp} \) is \( s^2 \), so smaller \( K_{sp} \) means smaller \( s \). For 2:1 salts, \( K_{sp} = 4s^3 \), so the relationship is different. But in this case, even though \( Ag_2CO_3 \) has a larger \( K_{sp} \) than \( AgBr \), its solubility is higher than \( AgCl \) (which has a larger \( K_{sp} \) than \( Ag_2CO_3 \))? Wait, no, \( AgCl \)'s \( K_{sp} \) is \( 1.8e-10 \), which is larger than \( Ag_2CO_3 \)'s \( 8.0e-12 \). But because \( Ag_2CO_3 \) dissociates into more ions, its molar solubility is higher. So the key is that for salts with the same ion ratio, \( K_{sp} \) determines solubility; for different ratios, we need to calculate molar solubility.

So, summarizing:
- \( AgBr \) (1:1): \( s = \sqrt{5.4e-13} \approx 7.35e-7 \, M \)
- \( AgCl \) (1:1): \( s = \sqrt{1.8e-10} \approx 1.34e-5 \, M \)
- \( Ag_2CO_3 \) (2:1): \( s = (8.0e-12 / 4)^{1/3} \approx 1.26e-4 \, M \)

Order of solubility (least to most): \( AgBr < AgCl < Ag_2CO_3 \)

Answer:

b. \( \text{AgBr} < \text{AgCl} < \text{Ag}_2\text{CO}_3 \)