Question
calculating concentration of Pb2+ ions in saturated PbCl2 solution with 0.023 M chloride ions
Answers
Answered by
GPT-4o mini
To calculate the concentration of \( \text{Pb}^{2+} \) ions in a saturated solution of \( \text{PbCl}_2 \) (lead(II) chloride) given that the concentration of chloride ions \( [\text{Cl}^-] \) is 0.023 M, you can start with the solubility product constant \( K_{sp} \) for \( \text{PbCl}_2 \).
The dissociation of \( \text{PbCl}_2 \) in water can be represented as:
\[
\text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Cl}^- (aq)
\]
The equilibrium expression for the solubility product is:
\[
K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2
\]
For lead(II) chloride, the \( K_{sp} \) value at 25 °C is typically around \( 1.7 \times 10^{-5} \).
Let's denote the concentration of \( \text{Pb}^{2+} \) as \( s \). From the dissociation, we see that for every mole of \( \text{PbCl}_2 \) that dissolves, it produces one mole of \( \text{Pb}^{2+} \) and two moles of \( \text{Cl}^- \). Thus, at equilibrium:
- \( [\text{Pb}^{2+}] = s \)
- \( [\text{Cl}^-] = 0.023 \, \text{M} + 2s \)
However, since \( s \) (the concentration of \( \text{Pb}^{2+} \)) is expected to be very small compared to 0.023 M, we can simplify the expression and consider \( [\text{Cl}^-] \approx 0.023 \, \text{M} \) (we will neglect \( 2s \) for this approximation).
Now substituting into the \( K_{sp} \) expression:
\[
K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = s (0.023)^2
\]
Now, plug in the value of \( K_{sp} \):
\[
1.7 \times 10^{-5} = s (0.023)^2
\]
Calculating \( (0.023)^2 \):
\[
(0.023)^2 = 0.000529
\]
Substituting this back into the equation:
\[
1.7 \times 10^{-5} = s \times 0.000529
\]
Now solve for \( s \):
\[
s = \frac{1.7 \times 10^{-5}}{0.000529} \approx 0.0322 \, \text{M}
\]
Thus, the concentration of \( \text{Pb}^{2+} \) ions in the saturated solution of \( \text{PbCl}_2 \) with 0.023 M chloride ions is approximately **0.0322 M**.
The dissociation of \( \text{PbCl}_2 \) in water can be represented as:
\[
\text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Cl}^- (aq)
\]
The equilibrium expression for the solubility product is:
\[
K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2
\]
For lead(II) chloride, the \( K_{sp} \) value at 25 °C is typically around \( 1.7 \times 10^{-5} \).
Let's denote the concentration of \( \text{Pb}^{2+} \) as \( s \). From the dissociation, we see that for every mole of \( \text{PbCl}_2 \) that dissolves, it produces one mole of \( \text{Pb}^{2+} \) and two moles of \( \text{Cl}^- \). Thus, at equilibrium:
- \( [\text{Pb}^{2+}] = s \)
- \( [\text{Cl}^-] = 0.023 \, \text{M} + 2s \)
However, since \( s \) (the concentration of \( \text{Pb}^{2+} \)) is expected to be very small compared to 0.023 M, we can simplify the expression and consider \( [\text{Cl}^-] \approx 0.023 \, \text{M} \) (we will neglect \( 2s \) for this approximation).
Now substituting into the \( K_{sp} \) expression:
\[
K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = s (0.023)^2
\]
Now, plug in the value of \( K_{sp} \):
\[
1.7 \times 10^{-5} = s (0.023)^2
\]
Calculating \( (0.023)^2 \):
\[
(0.023)^2 = 0.000529
\]
Substituting this back into the equation:
\[
1.7 \times 10^{-5} = s \times 0.000529
\]
Now solve for \( s \):
\[
s = \frac{1.7 \times 10^{-5}}{0.000529} \approx 0.0322 \, \text{M}
\]
Thus, the concentration of \( \text{Pb}^{2+} \) ions in the saturated solution of \( \text{PbCl}_2 \) with 0.023 M chloride ions is approximately **0.0322 M**.