Question

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

Short answer

The molar solubility of $\mathrm{SrF}_{2}$ in water, ignoring the basic properties of $\mathrm{F}^{-}$, is \(1.61 \times 10^{-3}\,\text{M}\). When considering the basic properties of $\mathrm{F}^{-}$, the measured molar solubility would be greater than the calculated value. In a buffered solution at $\mathrm{pH}=2.00$, the molar solubility of $\mathrm{SrF}_{2}$ is approximately \(1.61 \times 10^{-3}\, \text{M}\).

Step-by-step solution

a. Molar solubility of \(\mathrm{SrF}_{2}\) in water ignoring basic properties of \(\mathrm{F}^{-}\)

To calculate the molar solubility, we need to write the solubility equation and use the solubility product constant (\(K_{sp}\)). The solubility equation is: \[ \mathrm{SrF}_{2} \rightarrow \mathrm{Sr}^{2+} + 2\mathrm{F}^{-} \] Let x be the molar solubility of \(\mathrm{SrF}_{2}\). Then, at equilibrium: \[ [\mathrm{Sr}^{2+}] =x \] and \[ [\mathrm{F}^{-}]=2x \] Now, we can use the \(K_{sp}\) expression to find x: \[ K_{sp}=[\mathrm{Sr}^{2+}][\mathrm{F}^{-}]^{2}=7.9 \times 10^{-10} \] Substitute the concentration expressions in terms of x and solve: \[K_{sp}=x(2x)^{2}=7.9 \times 10^{-10} \] \[4x^{3}=7.9 \times 10^{-10} \] \[x^{3}=\frac{7.9 \times 10^{-10}}{4} \] \[x=\sqrt[3]{\frac{7.9 \times 10^{-10}}{4}} \] \[x=1.61 \times 10^{-3} \] Thus, the molar solubility of \(\mathrm{SrF}_{2}\) in water, ignoring the basic properties of \(\mathrm{F}^{-}\), is \(1.61 \times 10^{-3}\,\text{M}\).

b. Comparison of calculated molar solubility with measured value considering basic properties of \(\mathrm{F}^{-}\)

If we consider the basic properties of \(\mathrm{F}^{-}\), this ion will react with water to form OH\(^-\) and HF: \[ \mathrm{F}^{-}(\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{HF}(\mathrm{aq}) + \mathrm{OH}^{-}(\mathrm{aq}) \] This reaction leads to the consumption of \(\mathrm{F}^{-}\) ions, effectively decreasing their concentration in the solution. As a result, the solubility product equilibrium will shift in the forward direction, leading to increased dissolution of more \(\mathrm{SrF}_{2}\). Therefore, the measured molar solubility of \(\mathrm{SrF}_{2}\) would be greater than the calculated value from part a when considering the basic properties of \(\mathrm{F}^{-}\).

c. Molar solubility of \(\mathrm{SrF}_{2}\) in a buffered solution at \(\mathrm{pH}=2.00\)

At a pH of 2.00, we can calculate the initial concentration of \(\mathrm{H}^{+}\) in the buffered solution: \[ [\mathrm{H}^{+}]=10^{-\mathrm{pH}}=10^{-2}=\frac{1}{100}\,\text{M} \] Considering the basic properties of \(\mathrm{F}^{-}\), it will react with \(\mathrm{H}^{+}\) to form \(\mathrm{HF}\): \[ \mathrm{F}^{-} + \mathrm{H}^{+} \rightleftharpoons \mathrm{HF} \] Using an ICE table to represent the change in concentrations: Initial: \(x\) \(x+10^{-2}\) 0 Change: -y --y +y Equilibrium: \(x-y\) \((x+10^{-2})-y\) y As we are given the \(K_{a}\) of \(\mathrm{HF}\), we can write the equilibrium expression and solve for y: \[K_{a}=[\mathrm{HF}][\mathrm{H}^{+}]/[\mathrm{F}^{-}]=7.2 \times 10^{-4} \] \[7.2 \times 10^{-4} =\frac{y\left[(x+10^{-2})-y\right]}{x-y} \] Since we have already determined the molar solubility of \(\mathrm{SrF}_{2}\) without considering the \(\mathrm{F}^{-}\) basic properties, we know that x = \(1.61 \times 10^{-3}\,\text{M}\). Given that \(K_{a}\) is much smaller than \(1\), we can assume that y << x. Therefore, we can approximate the equilibrium concentrations as follows: \[7.2 \times 10^{-4} = \frac{y(1.61 \times 10^{-3})}{1.61 \times 10^{-3} - y} \] Now, we can solve for y: \[y = 2.15 \times 10^{-6}\, \text{M}\] With y determined, we can find the equilibrium concentration of \(\mathrm{F}^{-}\): \[[\mathrm{F}^{-}] = x - y = 1.61 \times 10^{-3} - 2.15 \times 10^{-6} \approx 1.61 \times 10^{-3}\, \text{M} \] Therefore, the molar solubility of \(\mathrm{SrF}_{2}\) in the buffered solution at pH 2.00 is approximately \(1.61 \times 10^{-3}\, \text{M}\).

Key concepts

Solubility Product Constant (Ksp)

Understanding the concept of the solubility product constant is crucial for predicting and analyzing the solubility of ionic compounds in water. It represents a special type of equilibrium constant, denoted as Ksp, which applies to the dissolution of sparingly soluble salts. The Ksp value provides insight into how much of a salt can dissolve in a solution to reach a point of saturation.
When a salt like SrF2 dissociates in water, it forms Sr2+ and F- ions. The Ksp expression for SrF2 would look like this:
\[ K_{sp} = [Sr^{2+}][F^{-}]^2 \]
The brackets signify the molar concentration of ions at equilibrium. A higher Ksp value indicates a more soluble compound. To calculate the molar solubility, which is the maximum amount of salt that can dissolve in a solution, one would set up an equation using Ksp and the stoichiometry of the dissolution reaction. Simplification of this equation requires critical thinking to avoid common mistakes, such as neglecting square powers for ions produced in a 1:n ratio. Therefore, understanding Ksp not only helps in calculating molar solubility but also in understanding ionic interactions within a solution.

Equilibrium Concentration

Equilibrium concentration is a term that refers to the concentration of each reactant and product in a reaction mixture when the reaction has reached a state of balance and no further change in the concentration of reactants and products is observed. In the context of solubility, it pertains to the amount of ions in solution when the dissolving and precipitation processes of a salt occur at equal rates.
For instance, in the exercise with SrF2, equilibrium concentrations are designated with square brackets and written as [Sr2+] and [F-]. Initially, we consider the undissolved SrF2 as having a concentration of 'x', which upon reaching equilibrium, dissociates to provide x amount of Sr2+ ions and 2x amount of F- ions.
In cases where additional reactions influence the ion concentrations, such as the formation of HF in acidic conditions, an ICE (Initial, Change, Equilibrium) table can be utilized to visually aid in understanding the changes in concentrations from start to equilibrium. Understanding how to calculate and interpret equilibrium concentrations is essential for predicting the extent of a reaction and for calculating related parameters, like the solubility product constant.

Buffered Solution pH

The pH of a buffered solution is a measure of its acidity or basicity. Buffers resist changes in pH when small amounts of acid or base are added, making them important in many chemical and biological applications. The pH of a buffer is directly influenced by its components - typically a weak acid and its conjugate base - and is calculated using the Henderson-Hasselbalch equation.
When dealing with solubility calculations, like those of SrF2 in a buffered solution with a known pH, understanding buffer chemistry becomes indispensable. In the given exercise, the buffered solution with pH 2.00 implies a high concentration of H+ ions, which will interact with any F- ions present, effectively shifting the equilibrium. The presence of high H+ concentration competes with Sr2+ for F- ions, forming HF and thus, disturbing the initial solubility equilibrium.
The significance of buffered solution pH is seen in adjusting the molar solubility of a compound by controlling the pH of the solution. Through careful manipulation of the buffer pH, desired solubility properties can be achieved, which is extremely valuable in industrial processes and in maintaining biological systems.