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Answer :
The [tex]K_{sp}[/tex] for [tex]Ag_2SO_3[/tex] is calculated to be [tex]1.49 \times 10^{-14}[/tex] using its molar solubility in pure water. In a 0.0250 M [tex]AgNO_3[/tex] solution, the molar solubility of [tex]Ag_2SO_3[/tex] is [tex]2.38 \times 10^{-11} M[/tex] due to the common ion effect.
First, we determine the [tex]K_{sp}[/tex] for [tex]Ag_2SO_3[/tex]. The dissociation of [tex]Ag_2SO_3[/tex] in water can be represented as:
[tex]Ag_2SO_3(s) \leftrightharpoons 2Ag^+(aq) + SO_3^{2-}(aq)[/tex]
If the molar solubility (S) of [tex]Ag_2SO_3[/tex] is [tex]1.55 \times 10^{-5} M[/tex], then:
- [tex][Ag^+] = 2S = 2 \times 1.55 \times 10^{-5} M = 3.10 \times 10^{-5} M[/tex]
- [tex][SO_3^{2-}] = S = 1.55 \times 10^{-5} M[/tex]
The [tex]K_{sp}[/tex] expression for [tex]Ag_2SO_3[/tex] is:
- [tex]K_{sp} = [Ag^+]^2[SO_3^{2-}] = (3.10 \times 10^{-5})^2(1.55 \times 10^{-5})[/tex]
Calculating this gives:
- [tex]K_{sp} = (3.10 \times 10^{-5})^2 \times 1.55 \times 10^[-5} = 1.49 \times 10^{-14}[/tex]
Next, we'll calculate the molar solubility of [tex]Ag_2SO_3[/tex] in 0.0250 M [tex]AgNO_3[/tex]. Because of the common ion effect, the concentration of Ag⁺ ions already in the solution is 0.0250 M:
Assume the additional solubility of [tex]Ag_2SO_3[/tex] in the presence of Ag⁺ ions is S'.
Thus:
- [tex][Ag^+] = 0.0250 + 2S' \approx 0.0250[/tex] since S' is very small.
- [tex][SO_3^{2-}] = S'[/tex]
Then,
- [tex]K_{sp} = (0.0250)^2 \times S'[/tex]
- [tex]1.49 \times 10^{-14} = (0.0250)^2 \times S'[/tex]
- [tex]S' = 1.49 \times 10^{-14} / (0.0250)^2 = 2.38 \times 10^{-11} M[/tex]
Therefore, the molar solubility of [tex]Ag_2SO_3[/tex] in 0.0250 M [tex]AgNO_3[/tex] is [tex]2.38 \times 10^{-11} M[/tex].
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Rewritten by : Barada
To calculate the Ksp for Ag2SO3, we use the following equation:
Ag2SO3(s) ⇌ 2Ag+(aq) + SO3^(2-)(aq)
Ksp = [Ag+]^2[SO3^(2-)]
Given the molar solubility of Ag2SO3 in pure water, we can calculate the solubility product constant (Ksp) as follows:
1.55 x 10^-5 = (2x)^2(x)
where x is the molar solubility of Ag2SO3 in pure water and 2x is the molar concentration of Ag+ ions.
Solving for x, we get:
x = 4.01 x 10^-6 M
Therefore, the Ksp for Ag2SO3 is:
Ksp = (4.01 x 10^-6)^2(2x10^-5) = 3.22 x 10^-17
To determine the molar solubility of Ag2SO3 in 0.0250 M AgNO3, we need to consider the common ion effect. AgNO3 is a soluble salt that dissociates in water to produce Ag+ and NO3- ions. Since the Ag+ ion is a common ion with the one produced by Ag2SO3, it will decrease the solubility of Ag2SO3.
Using the ICE table, we can calculate the new molar solubility of Ag2SO3 in the presence of 0.0250 M Ag+ ions:
Ag2SO3(s) ⇌ 2Ag+(aq) + SO3^(2-)(aq)
I 0.0250 0 0
C -2x +2x +x
E 0.0250-2x 2x x
Ksp = [Ag+]^2[SO3^(2-)]
Ksp = (2x)^2(x)
Ksp = 4x^3
Qsp = [Ag+]^2[SO3^(2-)]
Qsp = (0.0250)^2(4x)
Since Qsp < Ksp, the reaction is not at equilibrium and more Ag2SO3 can dissolve. Therefore, we can assume that 2x << 0.0250 and approximate the expression for Qsp as:
Qsp ≈ (0.0250)^2(4x) = 2.5 x 10^-6
Now, we can use the relationship between Qsp and Ksp to calculate the new molar solubility of Ag2SO3:
Ksp = Qsp
4x^3 = 2.5 x 10^-6
x = 2.77 x 10^-5 M
Therefore, the molar solubility of Ag2SO3 in 0.0250 M AgNO3 is 2.77 x 10^-5 M.
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