Question

(a) Will \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{p} \mathrm{H}\) of a 0.050 M solution of \(\mathrm{CaCl}_{2}\) is adjusted to 8.0? (b) Will \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) precipitate when 100 mL of 0.050 M \(\mathrm{AgNO}_{3}\) is mixed with 10 mL of \(5.0 \times 10^{-2} \mathrm{MNa}_{2} \mathrm{SO}_{4}\) solution?

Short answer

(a) For \(\mathrm{Ca}(\mathrm{OH})_{2}\), the ionic product, Q is calculated as \(Q = [Ca^{2+}] × [OH^-]^{2}\). Using the given pH and concentration of \(\mathrm{CaCl}_{2}\), we find that \(Q = 1.6 × 10^{-5}\). Since \(Q = 1.6 × 10^{-5} > K_{sp} = 5.02 × 10^{-6}\), precipitation of \(\mathrm{Ca}(\mathrm{OH})_{2}\) will occur. (b) For \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\), the ionic product Q is calculated as \(Q = [Ag^{+}]^{2} × [SO_{4}^{2-}]\). Using the given concentrations and volumes, we find that \(Q = 1.56 × 10^{-5}\). Since \(Q = 1.56 × 10^{-5} > K_{sp} = 1.20 × 10^{-5}\), precipitation of \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) will occur.

Step-by-step solution

Part A: Calculate the concentration of hydroxide ions

First, we need to calculate the concentration of hydroxide ions ([OH⁻]) in the solution. Since we know the pH of the solution, we can use the formula: pOH = 14 - pH And then to find the concentration of hydroxide ions, we use the formula: [OH⁻] = 10^(-pOH)

Part A: Calculate the ionic product for calcium hydroxide

Next, we need to calculate the ionic product (Q) for \(\mathrm{Ca}(\mathrm{OH})_{2}\), which is given by the formula: Q = [Ca²⁺] [OH⁻]² Given that the concentration of \(\mathrm{CaCl}_{2}\) is 0.050 M, and each molecule of \(\mathrm{CaCl}_{2}\) produces one Ca²⁺ ion, we can assume that the concentration of Ca²⁺ ions in the solution is also 0.050 M.

Part A: Determine if calcium hydroxide will precipitate

Finally, we need to compare the ionic product (Q) with the solubility product constant (Ksp) for \(\mathrm{Ca}(\mathrm{OH})_{2}\). The Ksp for \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(5.02 \times 10^{-6}\). If Q > Ksp, then precipitation will occur.

Part B: Calculate the moles of silver and sulfate ions

In this part, we need to determine the moles of silver ions (Ag⁺) and sulfate ions (SO₄²⁻) in the solution when mixing \(\mathrm{AgNO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). We can calculate the moles of each ion using the formula: moles = concentration × volume

Part B: Calculate the concentration of silver and sulfate ions

Now, we need to determine the concentrations of silver (Ag⁺) and sulfate (SO₄²⁻) ions in the combined solution. Since we know the total volume of the mixture (110 mL), we can use the formula: concentration = moles / total volume

Part B: Calculate the ionic product for silver sulfate

Next, we will determine the ionic product (Q) for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\). The ionic product is given by the formula: Q = [Ag⁺]² [SO₄²⁻]

Part B: Determine if silver sulfate will precipitate

Finally, we need to compare the ionic product (Q) with the solubility product constant (Ksp) for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\). The Ksp for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) is \(1.20 \times 10^{-5}\). If Q > Ksp, then precipitation will occur.

Key concepts

Precipitation Reaction

Precipitation reactions are fascinating chemical processes where ions in solution combine to form an insoluble compound that falls out of the solution as a solid, called a precipitate. In the realm of chemistry, understanding these reactions is crucial because they allow us to predict whether certain salts will remain dissolved or form a solid under different conditions.

For instance, when calcium hydroxide is potentially formed by mixing a calcium chloride solution with hydroxide ions, the solubility product constant (Ksp) becomes the key player in determining the outcome. If the ionic product, calculated from the concentrations of the individual ions, is greater than the Ksp, the solution is supersaturated and precipitation occurs. It is an intriguing balance game between soluble and insoluble that relies on concentration and the inherent solubility of the salt in question.

Ionic Product

Delving into the ionic product is like peering into the waters of solution equilibrium. Think of the ionic product as a snapshot of the ion concentrations in a solution at any given moment. To provide an analogy, consider ionic product as the tension in a rope during a tug-of-war. It changes as the concentrations of the ions involved in the precipitation reaction shift. The product of the concentrations of the reacting ions raised to the power of their stoichiometric coefficients as per the balanced equation gives us this key term, often represented as (Q).

In the case of calcium hydroxide, we identify the ionic product by multiplying the concentration of calcium ions by the square of the concentration of hydroxide ions. If this product exceeds the compound's Ksp, the 'tug-of-war' is won by precipitation. Otherwise, the ions remain contentedly dissolved.

pH and pOH Calculations

Navigating pH and pOH calculations is comparable to operating a GPS for the acidity or basicity of a solution. The pH scale is a measure of how acidic or basic water-based solutions are, with values ranging from 0 (extremely acidic) to 14 (extremely basic). The pOH serves as the counterpart to pH, focusing on the hydroxide ion concentration.

The relationship between pH and pOH is beautifully symmetrical in that they always add up to 14 at 25°C. That's the GPS route we take to calculate the hydroxide ion concentration in our solution: by subtracting the pH from 14 to get the pOH, and then using the antilog function to find the hydroxide ion concentration. It is a dance of logarithms and antilogarithms that allows chemists to traverse the landscape of acidity.

Concentration Calculations

Concentration calculations are the bread and butter of a chemist's quantitative analysis. They provide insight into the 'crowdedness' of particles in a given volume of solution. Let’s interpret it as knowing how many guests are in a party based on the number of invitations sent out and knowing the size of the hall.

By multiplying the molarity (the number of moles per liter of solution) by the volume, we obtain the number of moles of the solute. And when solutions are mixed, determining the new concentration after the party has combined guests is as simple as dividing the total moles of solute by the new volume. In the context of the exercise, this becomes essential when mixing silver nitrate and sodium sulfate, where the final concentration distinctly influences whether a precipitate of silver sulfate will crash the party or not.