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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
c. The temperature changes.

Step by step solution

01

Option A: The solution is stirred.

Stirring a solution increases the rate at which the solute dissolves, but it does not change the total amount of solute that can dissolve in the solvent. Once the solute particles are uniformly distributed throughout the solvent, stirring will no longer have any effect on the solubility.
02

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

Grinding the solute into fine particles increases its surface area, which speeds up the rate at which it dissolves. However, like stirring, grinding does not change the total amount of solute that can dissolve in a given amount of solvent. As the solute dissolves, the rate at which it dissolves may decrease as the concentration in the solution approaches the maximum solubility.
03

Option C: The temperature changes.

Changing the temperature will affect the total amount of solute that can dissolve in a given amount of solvent. The solubility of most solid solutes in a liquid solvent increases with increasing temperature. This is because an increase in temperature increases the kinetic energy of the solvent molecules, allowing them to interact with the solute particles more effectively and thus dissolve more solute. Conversely, decreasing the temperature usually decreases the solubility of solid solutes in liquid solvents. In conclusion, the correct answer is: c. The temperature changes.

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Most popular questions from this chapter

If \(10.0 \mathrm{mL}\) of $2.0 \times 10^{-3} M \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\( is added to 10.0 \)\mathrm{mL}$ of a \(\mathrm{pH}=10.0 \mathrm{NaOH}\) solution, will a precipitate form?

The copper(l) ion forms a complex ion with \(\mathrm{CN}^{-}\) according to the following equation: $$\mathrm{Cu}^{+}(a q)+3 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q) \quad K=1.0 \times 10^{11}$$ a. Calculate the solubility of $\mathrm{CuBr}(s)\left(K_{\mathrm{sp}}=1.0 \times 10^{-5}\right)\( in \)1.0 \mathrm{L}\( of \)1.0 \mathrm{M} \mathrm{NaCN}$ . b. Calculate the concentration of \(\mathrm{Br}^{-}\) at equilibrium. c. Calculate the concentration of \(\mathrm{CN}^{-}\) at equilibrium.

Sodium chloride is listed in the solubility rules as a soluble compound. Therefore, the \(K_{\mathrm{sp}}\) value for \(\mathrm{NaCl}\) is infinite. Is this statement true or false? Explain.

The concentration of Mg \(^{2+}\) in seawater is 0.052\(M .\) At what pH will 99\(\%\) of the \(\mathrm{Mg}^{2+}\) be precipitated as the hydroxide salt? $\left[K_{\mathrm{sp}} \text { for } \mathrm{Mg}(\mathrm{OH})_{2}=8.9 \times 10^{-12} .\right]$

Sulfide precipitates are generally grouped as sulfides insoluble in acidic solution and sulfides insoluble in basic solution. Explain why there is a difference between the two groups of sulfide precipitates.
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