Question

a. Calculate the molar solubility of \(\operatorname{Sr} \mathrm{F}_{2}\) in water, ignoring the basic properties of \(\left.\mathrm{F}^{-} . \text {(For } \operatorname{Sr} \mathrm{F}_{2}, K_{\mathrm{sp}}=7.9 \times 10^{-10} .\right)\). b. Would the measured molar solubility of \(\operatorname{Sr} \mathrm{F}_{2}\) be greater than or less than the value calculated in part a? Explain. c. Calculate the molar solubility of \(\operatorname{Sr} \mathrm{F}_{2}\) in a solution buffered at \(\mathrm{pH}=2.00 .\left(K_{\mathrm{a}} \text { for } \mathrm{HF} \text { is } 7.2 \times 10^{-4} .\right)\).

Short answer

a. The molar solubility of SrF2 in water is approximately \(1.33 \times 10^{-3}\) mole/L. b. The measured molar solubility of SrF2 would be less than the calculated value in part a due to the presence of F- ions from other sources, which decrease the solubility of SrF2 according to Le Chatelier’s principle. c. The molar solubility of SrF2 in a pH 2.00 buffered solution is approximately \(2.56 \times 10^{-6}\) mole/L.

Step-by-step solution

Write the dissociation equation and Ksp expression

We first need to write the dissociation equation for SrF2 and then write the Ksp expression using the given Ksp value. Dissociation equation: \(SrF_{2(s)} \rightleftharpoons Sr^{2+}_{(aq)} + 2F^-_{(aq)}\) Ksp = \([Sr^{2+}][F^-]^2 = 7.9 \times 10^{-10}\)

Define molar solubility in terms of x

Let x mole/L be the molar solubility of SrF2 in water. When SrF2 dissociates, we get 1x mole/L of Sr2+ ions and 2x mole/L of F- ions. So, Ksp = \([(1x)][(2x)]^2\)

Solve for x, the molar solubility

Now, we can solve the expression for x and find the molar solubility of SrF2. \(7.9 \times 10^{-10} = (1x)(4x^2)\) \(x^3 = \frac{7.9 \times 10^{-10}}{4}\) \(x \approx 1.33 \times 10^{-3}\) So, the molar solubility of SrF2 in water is approximately \(1.33 \times 10^{-3}\) mole/L. Part b: Comparing calculated molar solubility with the measured value

Brief explanation

The measured molar solubility of SrF2 would be less than the calculated value in part a due to the presence of F- ions from other sources. These additional F- ions would cause an increase in the [F-] concentration, which in turn shifts the equilibrium to the left (according to Le Chatelier’s principle), decreasing the solubility of SrF2. Part c: Calculate molar solubility of SrF2 in a pH 2.00 buffered solution

Convert pH to [H+] concentration

The pH of the solution is given as 2.00. We can find the concentration of H+ ions by using the pH formula: pH = -log([H+]) \[2.00 = -\log{[H^+]}\] \[10^{-2} = [H^+]\]

Write the expression for Ka of HF

The equilibrium expression for the acid dissociation constant, Ka, of HF is: \(Ka = \frac{[H^+][F^-]}{[HF]} = 7.2 \times 10^{-4}\) Since the solution is buffered, we can assume [F-] is equal to [HF]. Therefore, \[7.2 \times 10^{-4} = \frac{([H^+])([F^-])}{([F^-])}\]

Solve for [F-]

Now, solve the equation for [F-], using the concentration of H+ ions that we calculated earlier: \[7.2 \times 10^{-4} = \frac{(10^{-2})([F^-])}{([F^-])}\] \[[F^-] \approx 7.2 \times 10^{-2}\]

Write the Ksp expression and solve for molar solubility

In this case, Ksp = \([Sr^{2+}][F^-]^2\). We have determined the [F-] concentration in a pH 2.00 buffered solution. Now we can solve for x (molar solubility of SrF2). \(7.9 \times 10^{-10} = (1x)(7.2 \times 10^{-2})^2\) \(x \approx 2.56 \times 10^{-6}\) So, the molar solubility of SrF2 in a pH 2.00 buffered solution is approximately \(2.56 \times 10^{-6}\) mole/L.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, or Ksp, is a special type of equilibrium constant that measures the saturation level of a solute in a solution. It is specifically applied to the dissolution of sparingly soluble ionic compounds in water. When ionic compounds dissolve in water, they do so by dissociating into their constituent ions. The extent to which a compound can dissolve is represented by its Ksp value.

In the problem at hand, when we consider the dissociation of strontium fluoride (SrF2), we can set up an equilibrium expression based on this process. The smaller the Ksp value, the lower the solubility of the compound in water. For SrF2, with a Ksp of 7.9 x 10^-10, this suggests a low degree of solubility. The formula for Ksp in the context of SrF2 is \(Ksp = [Sr^{2+}][F^-]^2\). By establishing a variable x to represent the molar solubility and linking it to the concentrations of the ions at equilibrium, we can algebraically solve for x, determining the solubility of SrF2 in pure water.

Equilibrium Constant (Ka) for HF

The acid dissociation constant, or Ka, provides insight into the strength of an acid in solution. For hydrofluoric acid (HF), Ka represents the equilibrium constant for its dissociation into hydrogen ions (H+) and fluoride ions (F-). A higher Ka value indicates a stronger acid, which means it dissociates more in solution. The provided Ka for HF is 7.2 x 10^-4, reflecting that HF is a weak acid because it does not completely dissociate in solution.

In buffer solutions with a specific pH, such as pH 2.00, the concentration of hydrogen ions is known and can be used to calculate fluoride ions concentration originating from the dissociation of HF. This information is crucial as it affects the solubility of salts like SrF2 in such solutions. Because HF dissociates differently depending on the pH of the solution, the fluoride ion concentration at equilibrium (and thus the solubility of SrF2) can vary significantly. By calculating the fluoride ion concentration from the Ka expression \(Ka = [H^+][F^-]/[HF]\), we can adjust the Ksp expression accordingly to find the new solubility of SrF2 in a solution buffered to a specific pH.

Dissociation of Ionic Compounds in Water

Ionic compounds dissolve in water by breaking down into their component ions, a process known as dissociation. This behavior is crucial for understanding solubility and the consequent formation of a saturated solution. The dissociation is reversible, leading to an equilibrium state where the solid phase and the dissolved ions coexist.

For example, SrF2 dissociates in water as follows: \( SrF_{2(s)} \rightleftharpoons Sr^{2+}_{(aq)} + 2F^-_{(aq)} \). The extent of dissociation is important when predicting the solubility of a compound in water and in solutions with varying pH levels. The presence of common ions, in this case from added HF, can also shift the equilibrium according to Le Chatelier’s principle, affecting solubility. Additionally, the solubility of an ionic compound is influenced by the pH of the solution, especially if one of the ions is basic or acidic, as in the case of fluoride ions from SrF2.