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The solubility of $Ag_2CrO_4$ is determined to be $7.8 \times 10^{-5} M$. Calculate the $K_{sp}$ for $Ag_2CrO_4$.
The solubility of $Ag_2CrO_4$ is determined to be $7.8 \times 10^{-5} M$. Calculate the $K_{sp}$ for $Ag_2CrO_4$.
$K_{sp} = 4(7.8 \times 10^{-5})^3 = 1.9 \times 10^{-12}$
A solution contains $0.020 M$ $Ba^{2+}$ ions. What concentration of $SO_4^{2-}$ ions is needed to begin precipitation of $BaSO_4$? ($K_{sp}$ for $BaSO_4 = 1.1 \times 10^{-10}$)
A solution contains $0.020 M$ $Ba^{2+}$ ions. What concentration of $SO_4^{2-}$ ions is needed to begin precipitation of $BaSO_4$? ($K_{sp}$ for $BaSO_4 = 1.1 \times 10^{-10}$)
$[SO_4^{2-}] = \frac{K_{sp}}{[Ba^{2+}]} = \frac{1.1 \times 10^{-10}}{0.020} = 5.5 \times 10^{-9} M$
The $K_{sp}$ of $CaF_2$ is $3.9 \times 10^{-11}$. Calculate the molar solubility of $CaF_2$ in a solution containing $0.010 M$ $NaF$.
The $K_{sp}$ of $CaF_2$ is $3.9 \times 10^{-11}$. Calculate the molar solubility of $CaF_2$ in a solution containing $0.010 M$ $NaF$.
$K_{sp} = [Ca^{2+}][F^-]^2$; $3.9 \times 10^{-11} = (s)(2s + 0.010)^2$. Assume that $2s$ is small compared to 0.010, then, $3.9 \times 10^{-11} = s(0.010)^2$, $s = 3.9 \times 10^{-7} M$
Calculate the solubility of $AgCl$ in grams per liter in a $0.10 M$ $NaCl$ solution. ($K_{sp}$ for $AgCl = 1.8 \times 10^{-10}$, Molar mass of $AgCl = 143.32 g/mol$)
Calculate the solubility of $AgCl$ in grams per liter in a $0.10 M$ $NaCl$ solution. ($K_{sp}$ for $AgCl = 1.8 \times 10^{-10}$, Molar mass of $AgCl = 143.32 g/mol$)
The $K_{sp}$ of $Mg(OH)_2$ is $5.6 \times 10^{-12}$. At what pH will $Mg(OH)_2$ begin to precipitate from a solution with $[Mg^{2+}] = 0.010 M$?
The $K_{sp}$ of $Mg(OH)_2$ is $5.6 \times 10^{-12}$. At what pH will $Mg(OH)_2$ begin to precipitate from a solution with $[Mg^{2+}] = 0.010 M$?
A solution contains $0.010 M$ $Pb^{2+}$ and $0.010 M$ $Ba^{2+}$. If $Na_2SO_4$ is slowly added, which will precipitate first, $PbSO_4$ ( $K_{sp} = 1.8 \times 10^{-8}$) or $BaSO_4$ ($K_{sp} = 1.1 \times 10^{-10}$)?
A solution contains $0.010 M$ $Pb^{2+}$ and $0.010 M$ $Ba^{2+}$. If $Na_2SO_4$ is slowly added, which will precipitate first, $PbSO_4$ ( $K_{sp} = 1.8 \times 10^{-8}$) or $BaSO_4$ ($K_{sp} = 1.1 \times 10^{-10}$)?
For $PbCl_2$, $K_{sp} = 1.6 \times 10^{-5}$. Calculate the molar solubility of lead(II) chloride in 0.2 M $KCl$ solution.
For $PbCl_2$, $K_{sp} = 1.6 \times 10^{-5}$. Calculate the molar solubility of lead(II) chloride in 0.2 M $KCl$ solution.
Determine the $K_{sp}$ of $AgBr$ if its solubility is $5.7 \times 10^{-7} M$.
Determine the $K_{sp}$ of $AgBr$ if its solubility is $5.7 \times 10^{-7} M$.
Will a precipitate form when 10.0 mL of $2.0 \times 10^{-3} M$ $AgNO_3$ is mixed with 20.0 mL of $1.0 \times 10^{-3} M$ $NaCl$? ($K_{sp}$ of $AgCl = 1.8 \times 10^{-10}$)
Will a precipitate form when 10.0 mL of $2.0 \times 10^{-3} M$ $AgNO_3$ is mixed with 20.0 mL of $1.0 \times 10^{-3} M$ $NaCl$? ($K_{sp}$ of $AgCl = 1.8 \times 10^{-10}$)
Calculate the solubility of $AgCl$ in a solution that contains 0.050 M $NH_3$, given that $K_{sp}$ of $AgCl = 1.8 \times 10^{-10}$ and $K_f$ of $Ag(NH_3)_2^+ = 1.7 \times 10^7$.
Calculate the solubility of $AgCl$ in a solution that contains 0.050 M $NH_3$, given that $K_{sp}$ of $AgCl = 1.8 \times 10^{-10}$ and $K_f$ of $Ag(NH_3)_2^+ = 1.7 \times 10^7$.
Flashcards
What is Ksp?
What is Ksp?
The solubility product constant, representing the equilibrium constant for the dissolution of a sparingly soluble salt.
What is solubility?
What is solubility?
The concentration of a dissolved substance in a saturated solution.
What happens if Q > Ksp?
What happens if Q > Ksp?
The solid will precipitate out of the solution.
What happens if Q < Ksp?
What happens if Q < Ksp?
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How is solubility related to Ksp for AgCl?
How is solubility related to Ksp for AgCl?
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What is the common ion effect?
What is the common ion effect?
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What is the Ksp expression for Ag2CrO4?
What is the Ksp expression for Ag2CrO4?
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What happens to the molar solubility of AgCl if NaCl is added to the solution?
What happens to the molar solubility of AgCl if NaCl is added to the solution?
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Does temperature affect Ksp?
Does temperature affect Ksp?
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Study Notes
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Problem:
- The sparingly soluble salt, lead(II) iodide, PbI2, is added to 500.0 mL of water, forming a saturated solution at 25°C. The Ksp of PbI2 at 25°C is 7.1 x 10-9.
- Write the balanced chemical equation for the dissolution of PbI2(s) in water.
- Calculate the molar solubility of PbI2 in water at 25°C.
- Calculate the mass of PbI2(s) that dissolves in 500.0 mL of water.
- Calculate the [Pb2+] and [I-] in the solution.
- What is the ionic strength of the saturated solution?
- If 0.10 M lead(II) nitrate, Pb(NO3)2, is added to the saturated solution, what is the molar solubility of PbI2? (Assume the volume change is negligible.)
- Compared to the molar solubility of PbI2 in pure water, is the solubility higher, lower, or the same in the Pb(NO3)2 solution? Explain.
- The sparingly soluble salt, lead(II) iodide, PbI2, is added to 500.0 mL of water, forming a saturated solution at 25°C. The Ksp of PbI2 at 25°C is 7.1 x 10-9.
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Solution:
- Write the balanced chemical equation for the dissolution of PbI2(s) in water.
- PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)
- Calculate the molar solubility of PbI2 in water at 25°C.
- Ksp = [Pb2+][I-]2 = 7.1 x 10-9
- Let s = molar solubility of PbI2
- [Pb2+] = s
- [I-] = 2s
- Ksp = (s)(2s)2 = 4s3
- 4s3 = 7.1 x 10-9
- s3 = 1.775 x 10-9
- s = 1.21 x 10-3 M
- Calculate the mass of PbI2(s) that dissolves in 500.0 mL of water.
- Molar mass of PbI2 = 461.01 g/mol
- Mass = (1.21 x 10-3 mol/L) x (0.500 L) x (461.01 g/mol) = 0.279 g
- Calculate the [Pb2+] and [I-] in the solution.
- [Pb2+] = s = 1.21 x 10-3 M
- [I-] = 2s = 2.42 x 10-3 M
- What is the ionic strength of the saturated solution?
- Ionic strength (μ) = 1/2 Σ ci zi2
- μ = 1/2 ([Pb2+] (2+)2 + [I-] (1-)2)
- μ = 1/2 ((1.21 x 10-3 M)(4) + (2.42 x 10-3 M)(1))
- μ = 1/2 (4.84 x 10-3 + 2.42x10-3)
- μ = 3.63 x 10-3 M
- If 0.10 M lead(II) nitrate, Pb(NO3)2, is added to the saturated solution, what is the molar solubility of PbI2? (Assume the volume change is negligible.)
- PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)
- Initial [Pb2+] = 0.10 M
- Ksp = [Pb2+][I-]2 = 7.1 x 10-9
- Let s = molar solubility of PbI2
- [Pb2+] = 0.10 + s ≈ 0.10 M (since s is very small compared to 0.10)
- [I-] = 2s
- Ksp = (0.10)(2s)2 = 0.4s2
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- 4s2 = 7.1 x 10-9
- s2 = 1.775 x 10-8
- s = 1.33 x 10-4 M
- Compared to the molar solubility of PbI2 in pure water, is the solubility higher, lower, or the same in the Pb(NO3)2 solution? Explain.
- The solubility of PbI2 is lower in the Pb(NO3)2 solution. This is due to the common ion effect. The presence of Pb2+ ions from Pb(NO3)2 shifts the equilibrium of the dissolution of PbI2 to the left, decreasing the solubility of PbI2.
- Write the balanced chemical equation for the dissolution of PbI2(s) in water.
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