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Precipitation and Dissolution Quiz

12 questions chemistry Grades 9-12

The question sheet

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  1. Solubility equilibria are established when dissolution and precipitation occur at what rates?

    • Opposite rates
    • Increasing rates
    • Zero rates
    • Equal rates
    Reveal answer

    Answer: Equal rates

    Source evidence

    PDF page 834: Solubility equilibria are established when the dissolution and precipitation of a solute species occur at equal rates. These equilibria underlie many natural and technological processes, ranging from tooth decay to water purification. An understanding of the factors affecting compound solubility is, therefore, essential to the effective management of these processes. This section applies previously introduced equilibrium concepts and tools to systems involving dissolution and precipitation.

  2. The equilibrium constant for solubility equilibria is called the:

    • Reaction quotient
    • Molar solubility
    • Solubility product constant
    • Common ion constant
    Reveal answer

    Answer: Solubility product constant

    Source evidence

    PDF page 834: The equilibrium constant for solubility equilibria such as this one is called the solubility product constant, Ksp, in this case + − + − AgCl(s) ⇌ Ag (aq) + Cl (aq) Ksp = [Ag (aq)][Cl (aq)] Recall that only gases and solutes are represented in equilibrium constant expressions, so the Ksp does not include a term for the undissolved AgCl. A listing of solubility product constants for several sparingly soluble compounds is provided in Appendix J.

  3. Why does the Ksp expression NOT include a term for undissolved AgCl?

    • AgCl is a gas
    • Its concentration is zero
    • Solids are always constant
    • Only gases and solutes appear
    Reveal answer

    Answer: Only gases and solutes appear

    Source evidence

    PDF page 834: The equilibrium constant for solubility equilibria such as this one is called the solubility product constant, Ksp, in this case + − + − AgCl(s) ⇌ Ag (aq) + Cl (aq) Ksp = [Ag (aq)][Cl (aq)] Recall that only gases and solutes are represented in equilibrium constant expressions, so the Ksp does not include a term for the undissolved AgCl. A listing of solubility product constants for several sparingly soluble compounds is provided in Appendix J.

  4. A saturated solution added excess solute forms what kind of mixture?

    • A supersaturated gas
    • A pure compound
    • A homogeneous mixture
    • A heterogeneous mixture
    Reveal answer

    Answer: A heterogeneous mixture

    Source evidence

    PDF page 834: Recall from the chapter on solutions that the solubility of a substance can vary from essentially zero (insoluble or sparingly soluble) to infinity (miscible). A solute with finite solubility can yield a saturated solution when it is added to a solvent in an amount exceeding its solubility, resulting in a heterogeneous mixture of the saturated solution and the excess, undissolved solute. For example, a saturated solution of silver chloride is one in which the equilibrium shown below has been established. dissolution + − AgCl(s) Ag (aq) + Cl (aq) ⇌ precipitation + –

  5. For AgCl, the solubility product expression Ksp equals:

    • [AgCl]
    • [Ag+][Cl-]²
    • [Ag+]+[Cl-]
    • [Ag+][Cl-]
    Reveal answer

    Answer: [Ag+][Cl-]

    Source evidence

    PDF page 834: The equilibrium constant for solubility equilibria such as this one is called the solubility product constant, Ksp, in this case + − + − AgCl(s) ⇌ Ag (aq) + Cl (aq) Ksp = [Ag (aq)][Cl (aq)] Recall that only gases and solutes are represented in equilibrium constant expressions, so the Ksp does not include a term for the undissolved AgCl. A listing of solubility product constants for several sparingly soluble compounds is provided in Appendix J.

  6. For Mg(OH)2 dissolution, how many hydroxide ions are produced per formula unit?

    • One
    • Two
    • Four
    • Three
    Reveal answer

    Answer: Two

    Source evidence

    PDF page 835: (c) Mg(OH) (s) ⇌ Mg (aq) + 2OH (aq)

  7. In a CaF2 solution, the fluoride ion molarity equals what relative to calcium?

    • Three times its calcium molarity
    • Equal to its calcium molarity
    • Twice its calcium molarity
    • Half its calcium molarity
    Reveal answer

    Answer: Twice its calcium molarity

    Source evidence

    PDF page 836: of fluorite? Solution According to the stoichiometry of the dissolution equation, the fluoride ion molarity of a CaF2 solution is equal to twice its calcium ion molarity: 2 + − − −4 −4

  8. Comparing Qsp to Ksp predicts what about a solution?

    • The direction of the reaction
    • The number of ions
    • Its temperature
    • Its molar mass
    Reveal answer

    Answer: The direction of the reaction

    Source evidence

    PDF page 840: It is important to realize that this equilibrium is established in any aqueous solution containing Ca and CO3 ions, not just in a solution formed by saturating water with calcium carbonate. Consider, for example, mixing aqueous solutions of the soluble compounds sodium carbonate and calcium nitrate. If the concentrations of calcium and carbonate ions in the mixture do not yield a reaction quotient, Qsp, that exceeds the solubility product, Ksp, then no precipitation will occur. If the ion concentrations yield a reaction quotient greater than the solubility product, then precipitation will occur, lowering those concentrations until equilibrium is established (Qsp = Ksp). The comparison of Qsp to Ksp to predict precipitation is an example of the general approach to predicting the direction of a reaction first introduced in the chapter on equilibrium. For the specific case of solubility equilibria:

  9. If Qsp is greater than Ksp, what occurs?

    • Dissolution increases
    • Precipitation will occur
    • The solid disappears
    • No change
    Reveal answer

    Answer: Precipitation will occur

    Source evidence

    PDF page 841: Qsp < Ksp: the reaction proceeds in the forward direction (solution is not saturated; no precipitation observed) Qsp > Ksp: the reaction proceeds in the reverse direction (solution is supersaturated; precipitation will occur) This predictive strategy and related calculations are demonstrated in the next few example exercises.

  10. Why does very little barium sulfate dissolve when ingested for imaging?

    • It is a gas
    • It reacts with acid
    • It has a very small Ksp
    • It has a very large Ksp
    Reveal answer

    Answer: It has a very small Ksp

    Source evidence

    PDF page 839: barium sulfate, a chalky powder, is ingested by the patient. Since the Ksp of barium sulfate is 2.3 × 10 , very little of it dissolves as it coats the lining of the patient’s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure 15.4).

  11. The technique using controlled counter-ion addition to remove ions is called:

    • Recarbonation
    • Selective precipitation
    • Molar solubility
    • Common ion dilution
    Reveal answer

    Answer: Selective precipitation

    Source evidence

    PDF page 843: In solutions containing two or more ions that may form insoluble compounds with the same counter ion, an experimental strategy called selective precipitation may be used to remove individual ions from solution. By increasing the counter ion concentration in a controlled manner, ions in solution may be precipitated individually, assuming their compound solubilities are adequately different. In solutions with equal concentrations of target ions,

  12. In selective precipitation with equal ion concentrations, which precipitates first?

    • The least soluble compound
    • All simultaneously
    • The most soluble compound
    • The heaviest compound
    Reveal answer

    Answer: The least soluble compound

    Source evidence

    PDF page 844: the ion forming the least soluble compound will precipitate first (at the lowest concentration of counter ion), with the other ions subsequently precipitating as their compound’s solubilities are reached. As an illustration of this technique, the next example exercise describes separation of a two halide ions via precipitation of one as a silver salt.

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