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Chapter 7: Problem 80

How many times solubility of \(\mathrm{CaF}_{2}\) is decreased in \(4 \times 10^{-3} \mathrm{M}-\mathrm{KF}(\mathrm{aq})\) solution as compared to pure water at \(25^{\circ} \mathrm{C}\). Given: \(K_{\text {sp }}\left(\mathrm{CaF}_{2}\right)=3.2 \times 10^{-11}\) (a) 50 (b) 100 (c) 500 (d) 1000

Short Answer

Expert verified
The solubility of CaF2 is decreased by a factor of 100 in the presence of 4 × 10−3 M-KF(aq) solution compared to pure water.

Step by step solution

01

Write out the solubility product expression for CaF2

For the dissolution of CaF2, which is represented by the equation CaF2(s) ⇄ Ca^2+(aq) + 2F^-(aq),the solubility product expression is K_{sp} = [Ca^2+][F^-]^2.
02

Determine the ionic product in KF solution

In a solution of KF, F^- ions are provided by KF and will not affect the concentration of Ca^2+. The concentration of F^- ions due to KF is 4 × 10^-3 M. Thus, the ionic product in KF solution becomes[Ca^2+][4 × 10^-3]^2.
03

Calculate the decrease in solubility

Let the solubility of CaF2 in pure water be 's'. In pure water, the expression for Ksp is 3.2 × 10^-11 = [Ca^2+]_pure[s]^2. Now, in KF solution, [Ca^2+]_KF = [Ca^2+]_pure, because the presence of KF does not add or remove Ca^2+ ions.Since [F^-] is increased due to the KF, we can express this new solubility as 's / n', where 'n' is the factor by which the solubility has decreased. The new product becomes [Ca^2+](s/n)[4 × 10^-3]^2. This must also equal the constant Ksp.Equating the two expressions for Ksp we get:(s/n)^3(4 × 10^-3)^2 = s^2Solving for 'n' gives us the factor by which the solubility has decreased.
04

Rearrange and solve for 'n'

Rearranging the equation from the previous step to solve for 'n' gives usn^3 = (s^2)/((s/n)^3(4 × 10^-3)^2).Substitute in the value of Ksp for s^2:n^3 = (3.2 × 10^-11)/((4 × 10^-3)^2).Now, solve for 'n' to find the factor decrease in solubility.
05

Calculate 'n' and find the answer

Calculation:n^3 = (3.2 × 10^-11)/((4 × 10^-3)^2)n^3 = (3.2 × 10^-11)/(1.6 × 10^-5)n^3 = 2 × 10^4n = 2 × 10^4Taking the cube root of both sides, we have n = 2 × 10^{2 / 3}Therefore, n ≈ 100, which is the factor by which the solubility has decreased.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility in Chemistry
Understanding solubility in chemistry is essential for exploring how substances interact in solutions. Solubility refers to the maximum amount of a solute that can dissolve in a solvent at a given temperature and pressure to produce a saturated solution. Beyond this point, any additional solute will remain undissolved forming a precipitate.

In pure water, the solubility of ionic compounds is solely influenced by the solubility product constant, denoted as Ksp. However, in a solution containing a common ion, like KF in the given problem, the solubility of an ionic compound like CaF2 is reduced. This phenomena is known as the common ion effect. It dictates that the presence of a common ion in solution decreases the solubility of a solute that produces that same ion upon dissociation.

In our exercise, potassium fluoride (KF) provides additional fluoride ions (F^-), which are also produced by the dissociation of calcium fluoride (CaF2). By Le Chatelier's principle, the added fluoride ions shift the equilibrium towards the solid CaF2, reducing its solubility in the solution.
Ionic Product
The ionic product, often used interchangeably with the term 'reaction quotient' (Q), plays a critical role in assessing whether a precipitate will form in a solution. It represents the product of the concentrations of the ions in solution at any point in time, not just at equilibrium.

When the ionic product is compared to the solubility product constant, Ksp, it can provide valuable insights. If the ionic product is less than Ksp, the solution is unsaturated and more solute can dissolve. If it equals Ksp, the solution is at equilibrium and saturated. Lastly, if the ionic product exceeds Ksp, the solution is supersaturated, and a precipitate can form to reestablish equilibrium.

In the context of our problem, the KF solution introduces fluoride ions that increase the ionic product compared to the baseline scenario with pure water. Therefore, to maintain equilibrium with the unchanged Ksp, the concentration of dissolved CaF2 must decrease. This effectively lowers the solubility of CaF2 in the presence of KF.
Ksp Calculation
Calculation of the solubility product constant, or Ksp, is essential for understanding the solubility of sparingly soluble salts. Ksp is a temperature-dependent value and provides a quantitative measure of a salt's solubility in water.

In the calculation of solubility from Ksp, the concentrations of ions at equilibrium are raised to the power of their stoichiometric coefficients and then multiplied. For CaF2, this is represented by Ksp = [Ca^2+][F^-]^2. Calculating solubility in different scenarios often involves setting up an expression for Ksp in terms of the solubility (s) or the solubility divided by a factor (s/n), and then solving for the desired variable.

In the problem given, with additional fluoride ions from KF, the solubility of CaF2 in the KF solution—expressed as s/n—is calculated by equating the modified Ksp expression with the original Ksp value. Thus, by substituting and rearranging, we ultimately find factor 'n,' which indicates the decrease in solubility due to the presence of excess fluoride ions.

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Most popular questions from this chapter

The minimum mass of NaBr which should be added in \(200 \mathrm{ml}\) of \(0.0004 \mathrm{M}-\mathrm{AgNO}_{3}\) solution just to start the precipitation of AgBr. \(K_{\mathrm{sp}}\) of \(\mathrm{AgBr}=4 \times 10^{-13} \cdot(\mathrm{Br}=80)\) (a) \(1.0 \times 10^{-9} \mathrm{~g}\) (b) \(2 \times 10^{-10} \mathrm{~g}\) (c) \(2.06 \times 10^{-8} \mathrm{~g}\) (d) \(1.03 \times 10^{-7} \mathrm{~g}\)

Small amount of freshly precipitated \(\begin{array}{llll}\text { magnesium } & \text { hydroxides } & \text { are } & \text { stirred }\end{array}\) vigorously in a buffer solution containing \(0.25 \mathrm{M}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(0.05 \mathrm{M}\) of \(\mathrm{NH}_{4} \mathrm{OH}\). \(\left[\mathrm{Mg}^{2}\right]\) in the resulting solution is \(\left(K_{\mathrm{b}}\right.\) for \(\mathrm{NH}_{4} \mathrm{OH}=2.0 \times 10^{-5}\) and \(K_{\mathrm{sp}}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) \(\left.=8.0 \times 10^{-12}\right)\) (a) \(4 \times 10^{-6} \mathrm{M}\) (b) \(2 \times 10^{-6} \mathrm{M}\) (c) \(0.5 \mathrm{M}\) (d) \(2.0 \mathrm{M}\)

The solubility of sparingly soluble salt \(\mathrm{A}_{3} \mathrm{~B}_{2}\) (molar mass \(=\) 'M' \(\mathrm{g} / \mathrm{mol}\) ) in water is 'x' g/L. The ratio of molar concentration of \(\mathrm{B}^{3}\) to the solubility product of the salt is (a) \(\frac{108 x^{5}}{M^{5}}\) (b) \(\frac{x^{4}}{108 M^{4}}\) (c) \(\frac{x^{4}}{54 M^{4}}\) (d) \(\frac{x^{3}}{27 M^{3}}\)

The concentration of \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) ion in a solution prepared by adding \(0.1\) mole of \(\mathrm{CH}_{3} \mathrm{COOAg}(\mathrm{s})\) in \(1 \mathrm{~L}\) of \(0.1 \mathrm{M}-\mathrm{HCl}\) solution is [Given: \(K_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=10^{-5}\); \(\left.K_{\mathrm{sp}}(\mathrm{AgCl})=10^{-10} ; K_{\mathrm{sp}}\left(\mathrm{CH}_{3} \mathrm{COOAg}\right)=10^{-8}\right]\) (a) \(10^{-3} \mathrm{M}\) (b) \(10^{-2} \mathrm{M}\) (c) \(10^{-1} \mathrm{M}\) (d) \(1 \mathrm{M}\)

The solubility of \(\mathrm{AgCN}\) in a buffer solution of \(\mathrm{pH}=3.0\) is \(\left(K_{\mathrm{sp}}\right.\) of \(\mathrm{AgCN}\) \(=1.2 \times 10^{-16} ; K_{\mathrm{a}}\) of \(\left.\mathrm{HCN}=4.8 \times 10^{-10}\right)\) (a) \(1.58 \times 10^{-5} \mathrm{M}\) (b) \(2.0 \times 10^{-5} \mathrm{M}\) (c) \(1.58 \times 10^{-4} \mathrm{M}\) (d) \(2.5 \times 10^{-9} \mathrm{M}\)
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