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Chapter 17: Problem 57

Consider a beaker containing a saturated solution of \(\mathrm{CaF}_{2}\) in equilibrium with undissolved \(\mathrm{CaF}_{2}(s) .\) (a) If solid \(\mathrm{CaCl}_{2}\) is added to this solution, will the amount of solid \(\mathrm{CaF}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Ca}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of \(\mathrm{F}^{-}\) ions in solution increase or decrease?

Short Answer

Expert verified
(a) The amount of solid \(\mathrm{CaF}_{2}\) will increase. (b) The concentration of \(\mathrm{Ca}^{2+}\) ions in the solution will decrease. (c) The concentration of \(\mathrm{F}^{-}\) ions in the solution will decrease.

Step by step solution

01

Identify the common ion

The first step is to identify which ion is common between the \(\mathrm{CaF}_{2}\) and \(\mathrm{CaCl}_{2}\). In this case, it is the calcium ion, \(\mathrm{Ca}^{2+}\).
02

Write the solubility product expression for \(\mathrm{CaF}_{2}\)

To determine the effect of adding \(\mathrm{CaCl}_{2}\), we need to examine the solubility product expression for \(\mathrm{CaF}_{2}\). The solubility product expression is given as follows: \(K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2\)
03

Consider the effect of adding \(\mathrm{CaCl}_{2}\) on the equilibrium

When solid \(\mathrm{CaCl}_{2}\) is added to the solution, it will dissociate into \(\mathrm{Ca}^{2+}\) and \(\mathrm{Cl}^{-}\) ions. This increases the concentration of \(\mathrm{Ca}^{2+}\) ions in the solution. Since the concentration of \(\mathrm{Ca}^{2+}\) ions has increased and the solubility product constant, \(K_{sp}\), must remain constant, the product of the concentrations on the right-hand side of the equation must also remain constant. This will lead to a shift in the equilibrium as dictated by Le Châtelier's principle.
04

Analyze the effect on the amount of solid \(\mathrm{CaF}_{2}\) in the beaker

As we mentioned earlier, an increase in the concentration of \(\mathrm{Ca}^{2+}\) ions will cause a shift in the equilibrium. According to Le Châtelier's principle, the equilibrium will shift to counteract the disturbance, in this case the increased concentration of the common ion (\(\mathrm{Ca}^{2+}\)). Thus, the equilibrium will shift to the left, favoring the formation of more solid \(\mathrm{CaF}_{2}\). So, the amount of solid \(\mathrm{CaF}_{2}\) at the bottom of the beaker will increase. Answer: (a) The amount of solid \(\mathrm{CaF}_{2}\) will increase.
05

Analyze the effect on the concentration of \(\mathrm{Ca}^{2+}\) ions in the solution

As the equilibrium shifts to the left to reduce the increased concentration of \(\mathrm{Ca}^{2+}\) ions, the concentration of \(\mathrm{Ca}^{2+}\) ions will decrease in order to reach equilibrium and maintain the solubility product constant, \(K_{sp}\). Answer: (b) The concentration of \(\mathrm{Ca}^{2+}\) ions in the solution will decrease.
06

Analyze the effect on the concentration of \(\mathrm{F}^{-}\) ions in the solution

With the shift in equilibrium to the left, the system will attempt to reduce the increased concentration of \(\mathrm{Ca}^{2+}\) ions by forming more solid \(\mathrm{CaF}_{2}\). This reaction consumes \(\mathrm{F}^{-}\) ions as well. As a result, the concentration of \(\mathrm{F}^{-}\) ions in the solution will decrease. Answer: (c) The concentration of \(\mathrm{F}^{-}\) ions in the solution will decrease.

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Most popular questions from this chapter

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid \(\mathrm{HA}\) and a base \(\mathrm{B}\) are mixed. The \(\mathrm{pH}\) of the resulting solution is \(9.2 .(\mathrm{a})\) Write the equilibrium equation and equilibrium-constant expression for the reaction between \(\mathrm{HA}\) and \(\mathrm{B}\). (b) If \(K_{a}\) for HA is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between HA and \(B\) ? (c) What is the value of \(K_{b}\) for B?

(a) Calculate the pH of a buffer that is \(0.105 \mathrm{M}\) in \(\mathrm{NaHCO}_{3}\) and \(0.125 \mathrm{M}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). (b) Calculate the \(\mathrm{pH}\) of a solution formed by mixing \(65 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{NaHCO}_{3}\) with \(75 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\)

The osmotic pressure of a saturated solution of strontium sulfate at \(25^{\circ} \mathrm{C}\) is 21 torr. What is the solubility product of this salt at \(25^{\circ} \mathrm{C} ?\)

A buffer contains a weak acid, HX, and its conjugate base. The weak acid has a \(\mathrm{p} K_{a}\) of \(4.5,\) and the buffer has a \(\mathrm{pH}\) of \(4.3 .\) Without doing a calculation, predict whether \([\mathrm{HX}]=\left[\mathrm{X}^{-}\right]\) \([\mathrm{HX}]>\left[\mathrm{X}^{-}\right],\) or \([\mathrm{HX}]<\left[\mathrm{X}^{-}\right] .\) Explain. \([\) Section 17.2\(]\)

(a) Will \(\mathrm{Co}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{pH}\) of a \(0.020 \mathrm{M}\) solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) is adjusted to \(8.5 ?\) (b) Will \(\mathrm{AgIO}_{3}\) precipitate when \(20 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{NaIO}_{3} ?\left(K_{s p}\right.\) of \(\mathrm{AgIO}_{3}\) is \(\left.3.1 \times 10^{-8} .\right)\)
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