Question

Use \(K_{\text { sp values from table }} 17.3\) to predict whether a precipitate will form when equal volumes of the following solutions are mixed. a. 0.10\(M \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.030 \(\mathrm{M} \mathrm{NaF}\) b. 0.25\(M \mathrm{K}_{2} \mathrm{SO}_{4}\) and 0.010 \(\mathrm{M} \mathrm{AgNO}_{3}\)

Short answer

In case a, mixing 0.10 M \(Pb(NO_3)_2\) and 0.030 M \(NaF\) forms a precipitate of \(PbF_2\) since Q > K_sp (1.125 x 10^{-5} > 3.6 x 10^{-8}). In case b, mixing 0.25 M \(K_2SO_4\) and 0.010 M \(AgNO_3\) forms a precipitate of \(Ag_2SO_4\) since Q > K_sp (6.25 x 10^{-4} > 1.2 x 10^{-5}).

Step-by-step solution

Case a: Balanced chemical equation

: \(Pb(NO_3)_2\) + 2\(NaF\) \(\rightarrow\) \(PbF_2\) + 2\(NaNO_3\) The potential precipitate could be \(PbF_2\).

Case a: Calculate ion concentration

: When equal volumes of \(Pb(NO_3)_2\) and \(NaF\) solutions are mixed, concentrations are diluted by a factor of 2. \[ [Pb^{2+}]_{post-mix} = \frac{0.10}{2} = 0.05 M \] \[ [F^{-}]_{post-mix} = \frac{0.030}{2} = 0.015 M \]

Case a: Calculate Q

: Now we calculate the ion product Q: \[Q = [Pb^{2+}][F^-]^2\] \[Q = (0.05)(0.015)^2 = 1.125 \times 10^{-5}\]

Case a: Compare Q to K_sp

: From the table 17.3, the K_sp of \(PbF_2\) is 3.6 x 10^{-8}. Since Q > K_sp (1.125 x 10^{-5} > 3.6 x 10^{-8}), a precipitate of \(PbF_2\) will form.

Case b: Balanced chemical equation

: \(K_2SO_4\) + 2\(AgNO_3\) \(\rightarrow\) 2\(Ag_2SO_4\) + \(4KNO_3\) The potential precipitate could be \(Ag_2SO_4\).

Case b: Calculate ion concentration

: When equal volumes of \(K_2SO_4\) and \(AgNO_3\) solutions are mixed, concentrations are diluted by a factor of 2. \[ [SO_4^{2-}]_{post-mix} = \frac{0.25}{2} = 0.125 M \] \[ [Ag^{+}]_{post-mix} = \frac{0.010}{2} = 0.005 M \]

Case b: Calculate Q

: Now we calculate the ion product Q: \[Q = [Ag^+][SO_4^{2-}]\] \[Q = (0.005)(0.125) = 6.25 \times 10^{-4}\]

Case b: Compare Q to K_sp

: From the table 17.3, the K_sp of \(Ag_2SO_4\) is 1.2 x 10^{-5}. Since Q > K_sp (6.25 x 10^{-4} > 1.2 x 10^{-5}), a precipitate of \(Ag_2SO_4\) will form.

Key concepts

Solubility Product Constant (K_sp)

Understanding the solubility product constant, or K_sp, is crucial when predicting whether a chemical reaction will result in a precipitate. This value is unique for every sparingly soluble salt and represents the extent to which a compound can dissolve in water.

In simple words, it's a numerical expression of a substance's solubility equilibrium. When ions of a substance are in a saturated solution, they exist in dynamic equilibrium with the solid phase of the substance. The K_sp equation reflects the concentrations of these ions raised to the power of their coefficients in the balanced equation. For instance, if we have a compound AB that dissociates into A^+ and B^-, its K_sp expression would look like this: \[ K_{sp} = [A^+][B^-] \]

In the exercise provided, we use the K_sp values to determine if a mixture of different solutions will result in the formation of a precipitate. The key here is to compare the calculated product of the ion concentrations, known as the reaction quotient Q, with the known K_sp. If Q exceeds K_sp, it means that the solution is supersaturated, and a precipitate will form as the excess ions solidify to re-establish equilibrium.

Ion Concentration Calculation

Calculating ion concentrations involves figuring out the molarity of ions in a solution after mixing two solutions together. Molarity, symbolized by M, is a measure of the concentration of a solute in a solution, or how many moles of a given substance are in a liter of solution.

In our exercise, when we mix equal volumes of two solutions, the concentration of ions is halved because the volume of the solution doubles. For example, if we start with a 0.10 M solution of \(Pb(NO_3)_2\) and mix it with an equal volume of another solution, the concentration of \(Pb^{2+}\) ions becomes 0.05 M. It's important to correctly calculate these ion concentrations because they're used to determine the reaction quotient Q.

The process is quite straightforward but requires careful attention to ensure that the dilution of ions is accounted for properly. It's especially critical when solutions have different initial concentrations or if volumes are not equal. In such cases, the concentration of ions is calculated using the formula \(M_1V_1 = M_2V_2\), where M1 and V1 are the molarity and volume of the initial solution, and M2 and V2 are the molarity and volume after mixing.

Saturation Level Comparison

Comparing saturation levels in a solution allows us to predict whether a substance will remain dissolved or precipitate out. By comparing Q, the reaction quotient, which represents the current state of the solution, against K_sp, the solubility product constant, we can determine saturation levels.

If Q is less than K_sp, the solution is unsaturated, meaning more solute can dissolve. If Q equals K_sp, the solution is exactly saturated, and if Q is greater than K_sp, the solution is supersaturated, and a precipitate will form to reduce the ion concentration.

In the examples from the exercise, the calculated Q values were both greater than their respective K_sp values, indicating that precipitates will indeed form. This comparison is an invaluable tool in the prediction and understanding of precipitation reactions in various chemical processes, ensuring precise control over reaction conditions in lab settings or industrial applications.