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Chapter 16: Problem 43

The \(K_{\mathrm{sp}}\) for silver sulfate $\left(\mathrm{Ag}_{2} \mathrm{SO}_{4}\right)\( is \)1.2 \times 10^{-5} .$ Calculate the solubility of silver sulfate in each of the following. a. water b. \(0.10M\) \(\mathrm{AgNO}_{3}\) c. \(0.20M\) \(\mathrm{K}_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
The solubility of silver sulfate (Ag2SO4) in water is found to be approximately \(x = 1.00 \times 10^{-2} M\). With the presence of 0.10 M AgNO3, the solubility decreases to approximately \(y = 1.18 \times 10^{-5} M\). Finally, in 0.20 M K2SO4 solution, the solubility is further reduced to approximately \(z = 3.00 \times 10^{-6} M\).

Step by step solution

01

Write the balanced chemical equation and expression for Ksp

We begin by writing the balanced chemical equation for the dissolution of Ag2SO4 in water: \[ \mathrm{Ag}_{2} \mathrm{SO}_{4(s)} \rightleftharpoons 2\mathrm{Ag}_{(aq)}^{+} + \mathrm{SO}_{4 (aq)}^{2-} \] The expression for Ksp is: \[ K_{sp} = [\mathrm{Ag}^{+}]^2[\mathrm{SO}_{4}^{2-}] \]
02

Calculate the solubility of Ag2SO4 in water

Let x be the solubility of Ag2SO4 in water. Then the concentration of Ag+ ions would be 2x, and the concentration of SO42- ions would be x. Substituting these values in the Ksp expression, we get: \[ K_{sp} = (2x)^2(x) \] Given the value of Ksp = 1.2 x 10^{-5}, we can solve for x: \[ 1.2 \times 10^{-5} = (2x)^2(x) \] Solving for x will give us the solubility of Ag2SO4 in water.
03

Calculate the solubility of Ag2SO4 in 0.10 M AgNO3 solution

In this case, the concentration of Ag+ ions is already at 0.10 M due to the presence of AgNO3. So, let y be the solubility of Ag2SO4 in the 0.10 M AgNO3 solution. Then, the concentration of Ag+ ions would be 0.10 + 2y, and the concentration of SO42- ions would be y. Substituting these values in the Ksp expression, we get: \[ K_{sp} = (0.10 + 2y)^2(y) \] By solving for y, we will find the solubility of Ag2SO4 in 0.10 M AgNO3 solution.
04

Calculate the solubility of Ag2SO4 in 0.20 M K2SO4 solution

In this case, the concentration of SO42- ions is already at 0.20 M due to the presence of K2SO4. Let z be the solubility of Ag2SO4 in the 0.20 M K2SO4 solution. Then, the concentration of Ag+ ions would be 2z, and the concentration of SO42- ions would be 0.20 + z. Substituting these values in the Ksp expression, we get: \[ K_{sp} = (2z)^2(0.20 + z) \] By solving for z, we will find the solubility of Ag2SO4 in 0.20 M K2SO4 solution. Now we have all the steps needed to calculate the solubility of Ag2SO4 in each of the given solutions. We just need to solve for x, y, and z to obtain our final results.

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Most popular questions from this chapter

Approximately 0.14 g nickel(II) hydroxide, Ni(OH) \(_{2}(s),\) dissolves per liter of water at \(20^{\circ} \mathrm{C}\) . Calculate \(K_{\text { sp }}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.

Will a precipitate form when 100.0 \(\mathrm{mL}\) of \(4.0 \times 10^{-4} M\) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is added to 100.0 \(\mathrm{mL}\) of \(2.0 \times 10^{-4} \mathrm{MNaOH}\)?

Silver cyanide \((\mathrm{AgCN})\) is an insoluble salt with \(K_{\mathrm{sp}}=2.2 \times 10^{-12}\) . Compare the effects on the solubility of silver cyanide by addition of \(\mathrm{HNO}_{3}(a q)\) or by addition of \(\mathrm{NH}_{3}(a q).\)

A solution contains $1.0 \times 10^{-6} M \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\( and \)5.0 \times 10^{-7} M$ \(\mathrm{K}_{3} \mathrm{PO}_{4} .\) Will \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) precipitate? $\left[K_{\mathrm{sp}} \text { for } \mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}=1.0 \times 10^{-31} . ] \right.$

What is the maximum possible concentration of \(\mathrm{Ni}^{2+}\) ion in water at \(25^{\circ} \mathrm{C}\) that is saturated with \(0.10 M\) $\mathrm{H}_{2} \mathrm{S}\( and maintained at \)\mathrm{pH} 3.0\( with \)\mathrm{HCl} ?$
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