For a “common ion effect on solubility pogil answers” scenario, AgCl(s) has Ksp = 1.8×10^-10 at 25°C; what is the molar solubility of AgCl in pure water and in 0.10 M NaCl, and what percent decrease in solubility occurs?

The solubility is \(1.34\times10^{-5}\) M in pure water and approximately \(1.8\times10^{-9}\) M in 0.10 M NaCl, a decrease of about 99.99%.

Common Ion Effect on Solubility (POGIL-Style Practice)

Accepted answer Answer included

Problem

The phrase common ion effect on solubility pogil answers typically points to a solubility-equilibrium question where adding a common ion reduces the solubility of a sparingly soluble salt. Consider:

\(\mathrm{AgCl(s)}\) has \(\,K_{sp} = 1.8\times 10^{-10}\,\) at 25 °C. Find the molar solubility of \(\mathrm{AgCl}\) in (1) pure water and (2) a solution that is 0.10 M in \(\mathrm{NaCl}\). Then compute the percent decrease in solubility.

Direct “answer keys” to specific proprietary worksheets are not required to master this topic. Correct results follow from the same equilibrium setup: write \(K_{sp}\), define the equilibrium concentrations, and solve for the molar solubility \(s\).

Step-by-step solution

Step 1: Write the dissolution equilibrium and the K_sp expression

\(\mathrm{AgCl(s)}\) dissolves as:

\[ \mathrm{AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)} \]

\[ K_{sp} = [\mathrm{Ag^+}]\cdot[\mathrm{Cl^-}] \]

Step 2: Molar solubility in pure water

Let \(s_0\) be the molar solubility in pure water. Each mole of \(\mathrm{AgCl}\) produces one mole of \(\mathrm{Ag^+}\) and one mole of \(\mathrm{Cl^-}\), so:

\[ [\mathrm{Ag^+}] = s_0,\quad [\mathrm{Cl^-}] = s_0 \]

\[ K_{sp} = s_0^2 \quad \Rightarrow \quad s_0 = \sqrt{K_{sp}} = \sqrt{1.8\times 10^{-10}} = 1.34\times 10^{-5}\ \text{M} \]

Step 3: Molar solubility in 0.10 M NaCl (common ion present)

Let \(s\) be the molar solubility of \(\mathrm{AgCl}\) in 0.10 M \(\mathrm{NaCl}\). The solution already contains chloride from \(\mathrm{NaCl}\), so:

\[ [\mathrm{Ag^+}] = s,\quad [\mathrm{Cl^-}] = 0.10 + s \]

\[ K_{sp} = s(0.10 + s) \]

Solving gives:

\[ s^2 + 0.10s - K_{sp} = 0 \quad \Rightarrow \quad s = \frac{-0.10 + \sqrt{(0.10)^2 + 4\times K_{sp}}}{2} \]

Since \(4\times K_{sp}\) is negligible compared with \((0.10)^2\), the approximation \(0.10+s \approx 0.10\) is valid, so:

\[ s \approx \frac{K_{sp}}{0.10} = \frac{1.8\times 10^{-10}}{0.10} = 1.8\times 10^{-9}\ \text{M} \]

Step 4: Percent decrease in solubility

Percent decrease is computed relative to the pure-water solubility:

\[ \%\ \text{decrease} = \frac{s_0 - s}{s_0}\times 100\% \]

\[ \%\ \text{decrease} = \frac{1.34\times 10^{-5} - 1.8\times 10^{-9}}{1.34\times 10^{-5}}\times 100\% \approx 99.99\% \]

Summary table

Case	Equilibrium concentrations	K_sp equation	Molar solubility
Pure water	\([\mathrm{Ag^+}] = s_0,\ [\mathrm{Cl^-}] = s_0\)	\(K_{sp} = s_0^2\)	\(s_0 = 1.34\times 10^{-5}\ \text{M}\)
0.10 M NaCl	\([\mathrm{Ag^+}] = s,\ [\mathrm{Cl^-}] = 0.10 + s\)	\(K_{sp} = s(0.10+s)\)	\(s \approx 1.8\times 10^{-9}\ \text{M}\)
Change	\(\%\ \text{decrease} \approx 99.99\%\)

Visualization

The curve is generated from the equilibrium relationship \(s(C)=\frac{-C+\sqrt{C^2+4\times K_{sp}}}{2}\) with \(K_{sp}=1.8\times10^{-10}\). The two marked points correspond to pure water (\(C=0\)) and 0.10 M added chloride (\(C=0.10\)).

Final answers

Pure water: \(s_0 = 1.34\times 10^{-5}\ \text{M}\)
In 0.10 M NaCl: \(s \approx 1.8\times 10^{-9}\ \text{M}\)
Percent decrease: approximately \(99.99\%\)

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Problem

Step-by-step solution

Step 1: Write the dissolution equilibrium and the Ksp expression

Step 2: Molar solubility in pure water

Step 3: Molar solubility in 0.10 M NaCl (common ion present)

Step 4: Percent decrease in solubility

Summary table

Visualization

Final answers

Step 1: Write the dissolution equilibrium and the K_sp expression