Question

A solution is \(0.047 M\) in both \(\mathrm{NaF}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) Solid strontium nitrate, \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2},\) is added without changing the volume of the solution. (a) Which salt, \(\mathrm{SrCO}_{3}\) or \(\operatorname{SrF}_{2}\left(K_{\mathrm{sp}}=4.3 \times 10^{-9}\right),\) will precipitate first? (b) What is \(\left[\mathrm{Sr}^{2+}\right]\) when the salt in (a) first begins to precipitate?

Short answer

Answer: The salt SrCO₃ will precipitate first. The concentration of Sr²⁺ ions when SrCO₃ first begins to precipitate is \(\frac{K_{sp-SrCO3}}{0.047}\) M, where \(K_{sp-SrCO3}\) is the solubility product constant of SrCO₃.

Step-by-step solution

- Write the chemical equations for the formation of the salts

For both SrCO3 and SrF2 salts, write their dissociation chemical equations: For SrCO₃: \(SrCO_3 \rightleftharpoons Sr^{2+} + CO_3^{2-}\) For SrF₂: \(SrF_2 \rightleftharpoons Sr^{2+} + 2F^-\)

- Calculate the Ksp for SrCO3

We can calculate the solubility product constant (Ksp) for SrCO3 using Le Châtelier's principle. Since the carbonate ion (CO3²⁻) concentration in the solution is given (0.047 M), we can find the Ksp as follows: \(K_{sp} = [Sr^{2+}] [CO_3^{2-}]\) Since the concentrations of Sr2+ and CO3²⁻ are the same when the salt precipitates: \(K_{sp} = x * 0.047\) To find the Ksp for SrCO3, we need to compare this with Ksp for SrF₂, which is given as 4.3x10⁻⁹.

- Determine which salt precipitates first

We will now compare the reaction quotient (Q) with the Ksp values. If Q > Ksp, then the salt will precipitate. For SrCO₃: \(Q = [Sr^{2+}] [CO_3^{2-}] = x * 0.047\) For SrF₂: \( Q = [Sr^{2+}] [F^-]^2 = x * (0.047)^2\) Now, if the Q of SrCO₃ is greater than that of SrF₂, then SrCO₃ will precipitate first, and vice versa. Comparing the formulas we just wrote for Q: \(x * 0.047 > x * (0.047)^2\) Since x is the same in both cases, we can safely divide both sides by x: \(0.047 > 0.047^2\) This inequality holds true, meaning that SrCO₃ will precipitate first.

- Calculate the concentration of Sr2+

To find the concentration of Sr2+ ions when SrCO3 first begins to precipitate, we can use the Ksp equation for SrCO3: \(K_{sp} = [Sr^{2+}] [CO_3^{2-}]\) Now, we need to find the Ksp value for SrCO3. Since Q > Ksp and Q = x * 0.047 for SrCO3, we can find the Ksp for SrCO3: \(K_{sp-SrCO3}=x_{SrCO3} * 0.047\) Since [CO3²⁻] = 0.047 M and [Sr²⁺] = x, we can plug in the values into the Ksp equation for SrCO₃: \(K_{sp-SrCO3} = x * 0.047\) Solve for x: \(x = \frac{K_{sp-SrCO3}}{0.047}\) Now, we have the expression for the concentration of Sr²⁺ ions: \([Sr^{2+}] = \frac{K_{sp-SrCO3}}{0.047}\) The final answer of this problem will be the concentration of Sr²⁺ ions when SrCO3 first begins to precipitate.

Key concepts

Ksp

Understanding the Solubility Product Constant (Ksp) is key to predicting whether a salt will precipitate from a solution. It is a special type of equilibrium constant specific to the solubility of ionic compounds. To put it simply, Ksp is the product of the concentrations of the ions of a salt in a saturated solution, each raised to the power of their respective coefficients in the dissolution equation.

For example, the solubility product for strontium fluoride (SrF₂) with a Ksp of 4.3 × 10⁻⁹ can be expressed as:

[K_{sp} = [Sr^{2+}] [F^-]^2]
It represents the maximal product of the ion concentrations at which the solution is just saturated. Any increase above this value leads to precipitation, as the solution can no longer hold the dissolved salt.

Precipitation

Precipitation occurs when a substance solidifies from a solution, forming a solid, often referred to as a precipitate. This can happen when the product of the concentrations of the dissolved ions exceeds the solubility product constant (K_{sp}), resulting in the excess ions crystallizing out of the solution.

In the exercise, we analyze which of the two salts, SrCO₃ or SrF₂, will precipitate first. This decision is made by comparing the reaction quotient (Q) with the Ksp values. If Q exceeds Ksp for any of the salts, we expect it to start precipitating. The math behind this involves using concentrations of the ions already present in the solution, just as demonstrated through the solution steps for the provided exercise.

Chemical Equilibrium

Chemical equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction; hence, the concentrations of reactants and products remain constant over time. It's important to remember that reaching equilibrium does not mean the reactants and products are present in equal amounts, but rather their ratios no longer change.

In the context of soluble salts and Ksp, the equilibrium point is where the ionic components of a dissolved solid are in a dynamic balance with the solid particles. When we dissolve something like SrCO₃, it dissociates into Sr²⁺ and CO₃²⁻ ions. At equilibrium, the rate at which SrCO₃ forms from these ions is the same as the rate at which it dissolves back into them. And it's this delicate balance that guides the precipitation process.

Le Châtelier's Principle

Le Châtelier's Principle provides insight into how a system at equilibrium responds to disturbances. If a change, such as the addition of more reactant or removal of a product, is introduced to a system in equilibrium, the system will adjust to counteract the change and re-establish equilibrium.

This adjustment could involve shifting the equilibrium position to favor the forward or reverse reaction. In the exercise, by adding Sr(NO₃)₂, we introduce additional Sr²⁺ ions into the system. According to Le Châtelier's Principle, the presence of extra Sr²⁺ will shift the equilibrium towards the precipitate formation, hence leaning towards the side that favorably reduces the concentration of Sr²⁺, which in this case results in the precipitation of the less soluble salt.