Question

To \(100 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{AgNO}_{3}\) solution, solid \(\mathrm{K}_{2} \mathrm{SO}_{4}\) is added. The concentration of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) that shows the precipitation is \(\left(K_{s p}\right.\) for \(\left.\mathrm{A}_{\mathrm{g}_{2}} \mathrm{SO}_{4}=6.4 \times 10^{-5} \mathrm{M}\right)\) (1) \(0.1 \mathrm{M}\) (2) \(6.4 \times 10^{-3} \mathrm{M}\) (3) \(6.4 \times 10^{-7} \mathrm{M}\) (4) \(6.4 \times 10^{-5} \mathrm{M}\)

Short answer

Option 2: \(6.4 \times 10^{-3}\) M.

Step-by-step solution

- Identify the given data

The exercise provides the following data:- Volume of \text{AgNO}_3 = 100 \text{ mL} = 0.1 L- Concentration of \text{AgNO}_3 = 0.1 \text{ M}- Solubility product constant \text{K}_{sp} for \text{Ag}_2\text{SO}_4 = 6.4 \times 10^{-5} \text{ M}Ask for the concentration of \(\text{K}_2\text{SO}_4\) that causes precipitation.

- Write the precipitation reaction equation

The precipitation reaction is:$$2\text{Ag}^+ + \text{SO}_4^{2-} \rightarrow \text{Ag}_2\text{SO}_4$$

- Calculate the concentration of \text{Ag}^+ ions

Given concentration of \text{AgNO}_3 is 0.1 M and it dissociates completely in water:$$\text{AgNO}_3 \rightarrow \text{Ag}^+ + \text{NO}_3^-$$Thus, the concentration of \text{Ag}^+ is 0.1 M.

- Use the solubility product

The solubility product formula for \text{Ag}_2\text{SO}_4 is:$$K_{sp} = [\text{Ag}^+]^2[\text{SO}_4^{2-}]$$Given \(K_{sp} = 6.4 \times 10^{-5} \text{ M}\). With \([\text{Ag}^+] = 0.1 \text{ M}\), substitute these values in:$$6.4 \times 10^{-5} = (0.1)^2[\text{SO}_4^{2-}]$$

- Solve for sulfate ion concentration

Rearrange the equation to solve for \([\text{SO}_4^{2-}]\):$$[\text{SO}_4^{2-}] = \frac{6.4 \times 10^{-5}}{(0.1)^2}$$$$[\text{SO}_4^{2-}] = \frac{6.4 \times 10^{-5}}{0.01}$$$$[\text{SO}_4^{2-}] = 6.4 \times 10^{-3} \text{ M}$$

- Match the calculated concentration with the given options

The calculated concentration of \([\text{SO}_4^{2-}] = 6.4 \times 10^{-3}\) M matches option (2).Thus, the concentration of \(K_2SO_4\) that shows precipitation is \(6.4 \times 10^{-3}\) M.

Key concepts

Solubility Product Constant

The Solubility Product Constant, or \textit K_{sp}, is an equilibrium constant specific to the solubility of a compound. It helps determine how much of a solid dissolves in solution. For the compound \textit Ag_2SO_4, this constant is given as \textit K_{sp} = 6.4 \times 10^{-5} M. When \textit Ag_2SO_4 dissolves in water, it dissociates into its ions: $$\[\text{Ag}_2SO_4 \rightarrow 2\text{Ag}^+ + \text{SO}_4^{2-}\]$$ The expression for \text K_{sp} is: $$\[K_{sp} = [\text{Ag}^+]^2[\text{SO}_4^{2-}]\]$$ where \text[Ag^+] and \text[SO_4^{2-}] are the molar concentrations of the ions in solution. In this case, if we know the concentration of any ion, we can easily calculate the concentration of the other ion at equilibrium. This helps predict whether a precipitation reaction will occur, given a particular scenario.

Precipitation Reaction

A precipitation reaction is a chemical reaction where two solutions combine to form an insoluble solid called a precipitate. In the context of our exercise, when \text K_2SO_4 is added to \text AgNO_3, the ions in these compounds may form a precipitate if the conditions are right. The reaction of interest is: $$\[2\text{Ag}^+ + \text{SO}_4^{2-} \rightarrow \text{Ag}_2SO_4\]$$ When the product of the ion concentrations in the solution exceeds the \text K_{sp}, precipitation occurs. For silver sulfate (\text{Ag}_2SO_4), the formation of a solid precipitate depends on the ion product \text[Ag^+]^2[\text{SO}_4^{2-}] exceeding 6.4 × 10^{-5} M. The precipitation reaction is important in determining the solubility limit and will tell us if a solid will form when the concentrations of ions are increased by mixing solutions.

Ion Concentration Calculation

To calculate the ion concentration that causes precipitation, we need to know the \text K_{sp} and the initial ion concentrations. In the given problem, we start by identifying the dissociation of \text AgNO_3: $$\[\text{AgNO}_3 \rightarrow \text{Ag}^+ + \text{NO}_3^-\]$$ Given that \text[AgNO_3] = 0.1 M, the concentration of \text[Ag^+] is also 0.1 M. Using the \text K_{sp} value: $$\[K_{sp} = [\text{Ag}^+]^2[\text{SO}_4^{2-}]\]$$ $$\[6.4 \times 10^{-5} = (0.1)^2[\text{SO}_4^{2-}]\]$$ By rearranging, we solve for the sulfate ion concentration: $$\[6.4 \times 10^{-5} = 0.01 [\text{SO}_4^{2-}]\]$$ $$\[[\text{SO}_4^{2-}] = \frac{6.4 \times 10^{-5}}{0.01} = 6.4 \times 10^{-3} M\]$$ This means, when the concentration of \text[K_2SO_4] is 6.4 × 10^{-3} M, \text Ag_2SO_4 will precipitate from the solution.