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Chapter 16: Problem 1

Which of the following will affect the total amount of solute that can dissolve in a given amount of solvent? a. The solution is stirred. b. The solute is ground to fine particles before dissolving. c. The temperature changes.

Short Answer

Expert verified
Out of the given options, only option c (The temperature changes) will affect the total amount of solute that can dissolve in a given amount of solvent, as the solubility of most solid solutes increases with increasing temperature, whereas the solubility of gas solutes decreases with increasing temperature. Options a and b may speed up the dissolution process but do not change the solubility limit.

Step by step solution

01

Option a: The solution is stirred.

Stirring a solution can speed up the dissolution process, as it helps distribute the solute particles evenly throughout the solvent. However, stirring does not affect the total amount of solute that can ultimately dissolve in the solvent. The solubility limit of the solute remains constant.
02

Option b: The solute is ground to fine particles before dissolving.

Grinding the solute into fine particles increases the surface area of the solute that comes into contact with the solvent. This can speed up the dissolution process, but it does not change the total amount of solute that can dissolve in the solvent. The solubility limit remains the same.
03

Option c: The temperature changes.

Changing the temperature of the solvent can affect the total amount of solute that can dissolve in it. For most solid solutes, their solubility increases with increasing temperature. For gas solutes, however, their solubility decreases with increasing temperature. Therefore, the temperature change can directly affect the total amount of solute that can dissolve in a given amount of solvent. In conclusion, out of the given options, only option c (The temperature changes) will affect the total amount of solute that can dissolve in a given amount of solvent.

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Most popular questions from this chapter

A solution is prepared by adding \(0.10 \mathrm{~mol} \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) to \(0.50 \mathrm{~L}\) of \(3.0 \mathrm{M} \mathrm{NH}_{3} .\) Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text {owerall }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$5.5 \times 10^{\mathrm{x}}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$ for the overall reaction $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$

Calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)\) is relatively insoluble in water \(\left(K_{\mathrm{xp}}=\right.\) \(\left.2 \times 10^{-9}\right)\). However, calcium oxalate is more soluble in acidic solution. How much more soluble is calcium oxalate in \(0.10 \mathrm{M}\) \(\mathrm{H}^{+}\) than in pure water? In pure water, ignore the basic properties of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\)

A friend tells you: "The constant \(K_{\text {sp }}\) 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.

Silver chloride dissolves readily in \(2 \mathrm{M} \mathrm{NH}_{3}\), but is quite insoluble in \(2 \mathrm{M} \mathrm{NH}_{4} \mathrm{NO}_{3}\). Explain.

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