Chemistry, 19.03.2020 21:44 honeybaby675610

Silver sulfate dissolves in water according to the reaction: Ag2SO4(s) ∆ 2 Ag+(aq) + SO42-(aq) Kc = 1.1 * 10-5 at 298 K A 1.5-L solution contains 6.55 g of dissolved silver sulfate. If addi- tional solid silver sulfate is added to the solution, will it dissolve?

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Answer from: mayyyaaa

Silver sulfate dissolves in water according to the reaction: Ag2SO4(s) ∆ 2 Ag+(aq) + SO42-(aq) Kc = 1.1 * 10-5 at 298 K. A 1.5-L solution contains 6.55 g of dissolved silver sulfate. If additional solid silver sulfate is added to the solution, it will not dissolve because the ionic product is equal to the solubility product. When this scenario occurs, the solution in the reaction is saturated and no more solute can be dissolved.

From the given information, the equation for the reaction can be better expressed as:

$\mathbf{Ag_2SO_4 \to 2Ag^+_{(aq)} +SO_4^{2-} _{(aq)}}$

The solubility product is given as 1.1 × 10⁻⁵

Using an ICE table, we have:

$\mathbf{Ag_2SO_4 \to 2Ag^+_{(aq)} +SO_4^{2-} _{(aq)}}$

Initial - 0 0

Change - +2x +x

Equilibrium - 2x x

Now, the solubility product can be expressed as:

$\mathbf{K_{sp}}$ which is = $\mathbf{[Ag^+]^2 [SO_4^{2-}]}$

i.e.

$\mathbf{1.1 \times 10^{-5} = (2x)^2 \times (x)}$

$\mathbf{1.1 \times 10^{-5} = (2x)^3 }$

$\mathbf{ x= \sqrt[3]{\dfrac{1.1\times 10^{-5}}{2} } }$

x = 1.4 × 20⁻² M

Hence, the molar solubility of Ag2SO4(s) is 1.4 × 20⁻² M

We know that;

the concentration of a solution = number of moles/volumeNumber of moles = mass/molar mass

For Ag2SO4(s)

Number of moles = 6.55 g/311.8 g/molNumber of moles = 0.021 molesconcentration of Ag2SO4(s) = 0.021 moles/ 1.5 Lconcentration of Ag2SO4(s) = 0.014 M

Thus, from the given reaction:

$\mathbf{Ag_2SO_4 \to 2Ag^+_{(aq)} +SO_4^{2-} _{(aq)}}$

Since 1 mole of $\mathbf{Ag_2SO_4}$ result in the formation of 2 moles of $\mathbf{ 2Ag^+_{(aq)}}$

Then, 0.014 M = 2(0.014 M) for $\mathbf{ 2Ag^+_{(aq)}}$ = 0.0280 M for $\mathbf{ 2Ag^+_{(aq)}}$

Similarly, 1 mole of $\mathbf{Ag_2SO_4}$ results in the formation of 1 mole of $\mathbf{SO_4^{2-} _{(aq)}}$

∴

ionic product $\mathbf{K_{ip}}$ = (0.0280)²(0.0140)

= 1.1 × 10⁻⁵

Therefore, we can conclude that since $\mathbf{K_{sp}}$ is equal to $\mathbf{K_{ip}}$ , the addition of solid silver sulfate to the solution will not dissolve because the solution in the reaction is saturated and no more solute can be dissolved.

Learn more about solubility product here:

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Answer from: anastasiakonni1

additional solute will not dissolve.

Explanation:

First we must obtain the number of moles of Ag+ and SO4^2- ions present.

Number of moles of the compound = mass/ molar mass

Molar mass of compound= (2×108) + 32 + 4(16)= 216 + 32 + 64= 312gmol-1

Number of moles of compound= 6.55g/312gmol-1= 0.021moles

Concentration of Ag+ = (2×0.021)/1.5 =0.028M

Concentration of SO4^2- = 0.021/1.5= 0.014M

Qsp = [Ag+]^2[SO42-]= (0.028)^2 ( 0.014)=1.1 x 10^-5 = Ksp

Since Qsp=Ksp, additional solute will not dissolve.

Answer from: Quest

explanation:

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Answer from: Quest

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Silver sulfate dissolves in water according to the reaction: Ag2SO4(s) ∆ 2 Ag+(aq) + SO42-(aq) Kc =...

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