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Chapter 16: Problem 4

A friend tells you: "The constant \(K_{\mathrm{sp}}\) of a salt is called the solubility product constant and is calculated from the concentrations of ions in the solution. Thus, if salt A dissolves to a greater extent than salt \(\mathrm{B}\) , salt \(\mathrm{A}\) must have a higher \(K_{\mathrm{sp}}\) than salt \(\mathrm{B}\) ." Do you agree with your friend? Explain.

Short Answer

Expert verified
I agree with my friend's statement. If salt A dissolves to a greater extent than salt B, it means salt A has a higher solubility, which in turn leads to a higher concentration of ions in the solution at saturation. As the solubility product constant (Ksp) is calculated by multiplying the concentrations of the ions in a saturated solution, a higher solubility corresponds to a higher Ksp value. Therefore, salt A must have a higher Ksp than salt B.

Step by step solution

01

Understanding solubility product constant (Ksp)

Ksp, or solubility product constant, is a measure of a compound's solubility in a solvent, usually water. It is calculated by multiplying the concentrations of the ions in a saturated solution raised to the power of their stoichiometric coefficients. For a salt with a general formula AB, the dissolution process can be described as: \[AB_{(s)} \leftrightarrows A_{(aq)}^+ + B_{(aq)}^-\] Ksp can be represented as: \[K_{sp} = [A^+][B^-]\]
02

Solubility vs Ksp

A higher Ksp value means that the compound is more soluble in water. This is because a larger Ksp would require higher concentrations of ions in the solution for equilibrium to be achieved. Conversely, a lower Ksp value means that the compound is less soluble in water as the concentration of ions in the saturated solution would be lower.
03

Analyzing the friend's statement

The friend's statement suggests that if Salt A dissolves to a greater extent than Salt B, then Salt A must have a higher Ksp than salt B. Based on our understanding of Ksp and solubility, we can agree with the friend's statement. A higher solubility means that more ions of the salt will be present in the solution at saturation, and hence, will lead to a higher Ksp value.

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Most popular questions from this chapter

As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mixture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write equations for all the changes mentioned above. What conclusions can you draw concerning the sizes of the \(K_{\mathrm{sp}}\) values for \(\mathrm{AgCl}, \mathrm{AgBr},\) and \(\mathrm{AgI?}\)

In the presence of \(\mathrm{CN}^{-}, \mathrm{Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-} .\) The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) are $8.5 \times 10^{-40} \mathrm{M}\( and \)1.5 \times 10^{-3} M,\( respectively, in a \)0.11-M$ KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) . $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \qquad K_{\text { overall }}=?$$

When 100.0 \(\mathrm{mL}\) of 2.00 $\mathrm{M} \mathrm{Ce}\left(\mathrm{NO}_{3}\right)_{3}\( is added to 100.0 \)\mathrm{mL}$ of \(3.00 \mathrm{M} \mathrm{KIO}_{3},\) a precipitate of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}(s)\) forms. Calculate the equilibrium concentrations of \(\mathrm{Ce}^{3+}\) and \(\mathrm{IO}_{3}^{-}\) in this solution. $\left[K_{\mathrm{sp}} \text { for } \mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}=3.2 \times 10^{-10} .\right]$

List some ways one can increase the solubility of a salt in water.

Calculate the molar solubility of $\mathrm{Co}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=2.5 \times 10^{-43}.$
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