Question

0.20 mol nacl is added to 1.0 l of saturated \\(\text{pbcl}_2\\) solution. what is the \\(\text{pb}^{2+}\\) in the combined solution at 25 °c?

\\(\text{pbcl}_2\\) \\(k_{\text{sp}} = 1.7 \times 10^{-5}\\)

\\(\text{pb}^{2+} = ? \times 10^{?} \text{ m}\\)

Explanation:

Identify the chemical equilibrium and given values

We are given a saturated solution of lead(II) chloride, \(\text{PbCl}_2\), to which sodium chloride, \(\text{NaCl}\), is added.
The dissolution equilibrium for \(\text{PbCl}_2\) is:

$$\text{PbCl}_2(s) ightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq)$$

The solubility product constant expression is:

$$K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = 1.7 \times 10^{-5}$$

The volume of the solution is \(1.0\text{ L}\).
The amount of added \(\text{NaCl}\) is \(0.20\text{ mol}\).

Determine the concentration of chloride ions

Since \(\text{NaCl}\) is a strong electrolyte, it dissociates completely:

$$\text{NaCl}(aq) ightarrow \text{Na}^+(aq) + \text{Cl}^-(aq)$$

The concentration of \(\text{Cl}^-\) added from \(\text{NaCl}\) is:

$$[\text{Cl}^-]_{\text{added}} = \frac{0.20\text{ mol}}{1.0\text{ L}} = 0.20\text{ M}$$

Because of the Common Ion Effect, the solubility of \(\text{PbCl}_2\) is greatly suppressed.
The concentration of \(\text{Cl}^-\) coming from the dissolution of \(\text{PbCl}_2\) is \(2s\), where \(s = [\text{Pb}^{2+}]\).
Thus, the total equilibrium concentration of chloride is:

$$[\text{Cl}^-] = 0.20 + 2s$$

Apply the approximation for solubility

Since \(K_{sp} = 1.7 \times 10^{-5}\) is very small, the solubility \(s\) of \(\text{PbCl}_2\) in the presence of \(0.20\text{ M}\) \(\text{Cl}^-\) will be extremely small.
Therefore, we can make the approximation:

$$[\text{Cl}^-] = 0.20 + 2s \approx 0.20\text{ M}$$

Calculate the lead ion concentration

Substitute the approximated concentration of \(\text{Cl}^-\) into the \(K_{sp}\) expression:

$$K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2$$

$$1.7 \times 10^{-5} = [\text{Pb}^{2+}](0.20)^2$$

$$1.7 \times 10^{-5} = [\text{Pb}^{2+}](0.040)$$

$$[\text{Pb}^{2+}] = \frac{1.7 \times 10^{-5}}{0.040} = 4.25 \times 10^{-4}\text{ M}$$

Verify the approximation validity

Let's check if \(2s\) is indeed negligible compared to \(0.20\):

$$2s = 2(4.25 \times 10^{-4}) = 8.5 \times 10^{-4}\text{ M}$$

Comparing \(8.5 \times 10^{-4}\) to \(0.20\):

$$\frac{8.5 \times 10^{-4}}{0.20} \times 100\% \approx 0.425\%$$

Since this is well below \(5\%\), the approximation is highly accurate.
Thus, the concentration of \(\text{Pb}^{2+}\) is \(4.3 \times 10^{-4}\text{ M}\) (rounded to two significant figures).

Answer:

\(0.20\text{ mol NaCl}\) is added to \(1.0\text{ L}\) of saturated \(\text{PbCl}_2\) solution. What is the \([\text{Pb}^{2+}]\) in the combined solution at \(25\text{ }^\circ\text{C}\)?

\(\text{PbCl}_2\ \ K_{sp} = 1.7 \times 10^{-5}\)

\([\text{Pb}^{2+}] =\) <blank>\(4.3\)</blank> \(\times 10^{\text{ }}\)<blank>\(-4\)</blank> \(\text{M}\)