Consider an aqueous solution which is saturated with Ag2SO4. Given that Ksp of Ag2SO4 = 1.4 × 10–5. (a) Calculate [Ag+] in this saturated solution. Show your calculation. (b) What mass of Na2SO4 must be added to 0.500 L of the solution to decrease [Ag+] to 4.0 × 10–3 M? Show your calculation.
The Correct Answer and Explanation is:
(a) Calculation of [Ag+] in Saturated Solution
We are given that Ag2SO4 is saturated in an aqueous solution, and the solubility product (Ksp) of Ag2SO4 is:Ksp=1.4×10−5K_{\text{sp}} = 1.4 \times 10^{-5}Ksp=1.4×10−5
The dissociation of Ag2SO4 in water can be represented as:Ag2SO4(s)⇌2Ag+(aq)+SO42−(aq)\text{Ag}_2\text{SO}_4 (s) \rightleftharpoons 2\text{Ag}^+ (aq) + \text{SO}_4^{2-} (aq)Ag2SO4(s)⇌2Ag+(aq)+SO42−(aq)
Let the solubility of Ag2SO4 be denoted as sss mol/L. From the stoichiometry of the dissociation, we know that:
- The concentration of Ag+\text{Ag}^+Ag+ ions will be 2s2s2s mol/L (since 2 moles of Ag+ are produced for every mole of Ag2SO4 that dissolves).
- The concentration of SO42−\text{SO}_4^{2-}SO42− ions will be sss mol/L.
Now, we write the expression for the solubility product (Ksp):Ksp=[Ag+]2[SO42−]K_{\text{sp}} = [\text{Ag}^+]^2 [\text{SO}_4^{2-}]Ksp=[Ag+]2[SO42−]
Substitute the values for [Ag+]=2s[\text{Ag}^+] = 2s[Ag+]=2s and [SO42−]=s[\text{SO}_4^{2-}] = s[SO42−]=s:1.4×10−5=(2s)2⋅s1.4 \times 10^{-5} = (2s)^2 \cdot s1.4×10−5=(2s)2⋅s
Simplifying:1.4×10−5=4s31.4 \times 10^{-5} = 4s^31.4×10−5=4s3
Solve for sss:s3=1.4×10−54=3.5×10−6s^3 = \frac{1.4 \times 10^{-5}}{4} = 3.5 \times 10^{-6}s3=41.4×10−5=3.5×10−6s=3.5×10−63=1.52×10−2 Ms = \sqrt[3]{3.5 \times 10^{-6}} = 1.52 \times 10^{-2} \, \text{M}s=33.5×10−6=1.52×10−2M
Thus, the concentration of Ag+ ions in the saturated solution is:[Ag+]=2s=2×1.52×10−2=3.04×10−2 M[\text{Ag}^+] = 2s = 2 \times 1.52 \times 10^{-2} = 3.04 \times 10^{-2} \, \text{M}[Ag+]=2s=2×1.52×10−2=3.04×10−2M
(b) Mass of Na2SO4 to Decrease [Ag+] to 4.0 × 10–3 M
We are tasked with reducing the concentration of Ag+\text{Ag}^+Ag+ to 4.0×10−34.0 \times 10^{-3}4.0×10−3 M by adding Na2SO4. The addition of Na2SO4 will provide SO42−\text{SO}_4^{2-}SO42− ions, which will cause some of the Ag+\text{Ag}^+Ag+ ions to precipitate as Ag2SO4. The relationship between the concentrations of ions and the solubility product remains:Ksp=[Ag+]2[SO42−]K_{\text{sp}} = [\text{Ag}^+]^2 [\text{SO}_4^{2-}]Ksp=[Ag+]2[SO42−]
We want to reduce [Ag+][\text{Ag}^+][Ag+] to 4.0 × 10–3 M. Let’s assume the concentration of SO42−\text{SO}_4^{2-}SO42− added from Na2SO4 is xxx M.
At equilibrium, we know:1.4×10−5=(4.0×10−3)2⋅x1.4 \times 10^{-5} = (4.0 \times 10^{-3})^2 \cdot x1.4×10−5=(4.0×10−3)2⋅x
Solve for xxx:1.4×10−5=1.6×10−5⋅x1.4 \times 10^{-5} = 1.6 \times 10^{-5} \cdot x1.4×10−5=1.6×10−5⋅xx=1.4×10−51.6×10−5=0.875 Mx = \frac{1.4 \times 10^{-5}}{1.6 \times 10^{-5}} = 0.875 \, \text{M}x=1.6×10−51.4×10−5=0.875M
Now, we need to find the mass of Na2SO4 required to achieve this concentration of SO42−\text{SO}_4^{2-}SO42− in 0.500 L of solution. The number of moles of Na2SO4 required is:moles of Na2SO4=x×V=0.875 mol/L×0.500 L=0.4375 mol\text{moles of Na}_2\text{SO}_4 = x \times V = 0.875 \, \text{mol/L} \times 0.500 \, \text{L} = 0.4375 \, \text{mol}moles of Na2SO4=x×V=0.875mol/L×0.500L=0.4375mol
The molar mass of Na2SO4 is:Molar mass of Na2SO4=(2×23)+32+(4×16)=142 g/mol\text{Molar mass of Na}_2\text{SO}_4 = (2 \times 23) + 32 + (4 \times 16) = 142 \, \text{g/mol}Molar mass of Na2SO4=(2×23)+32+(4×16)=142g/mol
The mass of Na2SO4 required is:mass=moles×molar mass=0.4375 mol×142 g/mol=62.125 g\text{mass} = \text{moles} \times \text{molar mass} = 0.4375 \, \text{mol} \times 142 \, \text{g/mol} = 62.125 \, \text{g}mass=moles×molar mass=0.4375mol×142g/mol=62.125g
Thus, the mass of Na2SO4 that must be added is approximately:62.1 g\boxed{62.1 \, \text{g}}62.1g
