Question

To precipitate as \(\mathrm{Ag}_{2} \mathrm{S}(\mathrm{s}),\) all the \(\mathrm{Ag}^{+}\) present in \(338 \mathrm{mL}\) of a saturated solution of \(\mathrm{AgBrO}_{3}\) requires \(30.4 \mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) measured at \(23^{\circ} \mathrm{C}\) and \(748 \mathrm{mm} \mathrm{Hg} .\) What is \(K_{\mathrm{sp}}\) for \(\mathrm{AgBrO}_{3} ?\)

Short answer

The solubility product constant \(K_{sp}\) for \(AgBrO_{3}\) is \(0.00005041\).

Step-by-step solution

Determining Moles of \(Ag^{+}\)

The moles of \(\mathrm{Ag}^{+}\) can be determined by using the given volume of \(\mathrm{H}_{2}\mathrm{S}\) since the reaction between \(\mathrm{Ag}^{+}\) and \(\mathrm{H}_{2}\mathrm{S}\) forms \(\mathrm{Ag}_{2} \mathrm{S}\). Given the Ideal gas law we have the equation \(PV=nRT \Rightarrow n=PV/RT\). Here, \(P\) in Atmospheres will be \(748 \times 1 / 760 = 0.9842 \, atm\), \(V = 30.4 \, mL = 0.03 \, L\), \(R\) is the ideal gas constant = 0.0821 \(L.atm/K.mol\), and \(T\) is the temperature = 23 Celsius = 296.15 Kelvin. Substituting the values, we find number of moles of \(H_2S\) to be \(n = 0.0012 \, mol\). Since it reacts in a 1:2 ratio with \(Ag^{+}\), moles of \(Ag^{+}\) = 2 × moles of \(H_2S\) = 0.0024 mol.

Calculating Concentration of \(Ag^{+}\) and \(BrO_{3}^{-}\)

As per the reaction, one molecule of \(\mathrm{AgBrO}_{3}\) gives one ion each of \(Ag^{+}\) and \(BrO_{3}^{-}\). Therefore, concentration of \(Ag^{+}\) will be equal to the concentration of \(BrO_{3}^{-}\). The Volume of \(\mathrm{AgBrO}_{3}\) is 338 mL = 0.338 L. Hence, concentration = moles/volume = 0.0024 / 0.338 = 0.0071 M.

Calculating \(K_{sp}\) for \(AgBrO_{3}\)

The solubility product constant (\(K_{sp}\)) is given by: \(K_{sp}\) = [\(Ag^{+}\)] × [\(BrO_{3}^{-}\)] = (0.0071) × (0.0071) = 0.00005041.

Key concepts

Precipitation Reaction

Precipitation reactions play a crucial role in chemistry as they involve the formation of a solid, often referred to as a precipitate, from a solution. This occurs when two soluble salts react in solution to form one or more insoluble products. In the given exercise, the reaction between \(\mathrm{Ag}^{+}\) ions from a solution of \(\mathrm{AgBrO}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\) gas leads to the formation of the precipitate \(\mathrm{Ag}_{2} \mathrm{S}\).

The key to understanding precipitation reactions is recognizing that they depend heavily on the solubility rules for ionic compounds in water. When the product of the concentrations of the ions exceeds the solubility product constant (K_{sp}), a precipitate will form:

• A reaction occurs when the ionic product of the concentrations exceeds \(K_{sp}\).
• AgBrO_{3} and \(\mathrm{H}_{2} \mathrm{S}\) come together to form solid \(\mathrm{Ag}_{2} \mathrm{S}\).
• The formation of a precipitate removes ions from the solution, altering the equilibrium dynamics.

Thus, a precipitation reaction can help us understand the quantity and behavior of ions in a solution.

Ideal Gas Law

The Ideal Gas Law provides a mathematical relationship between pressure, volume, temperature, and the number of moles of a gas. It's essential for deriving quantities like the moles of gas used in reactions, as in the given problem with \(\mathrm{H}_{2}\mathrm{S}\). The formula to calculate these moles is \(PV=nRT\), where:

\(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant (0.0821 \(L.atm/K.mol\)), and \(T\) is temperature in Kelvin.

How to Use the Ideal Gas Law

To solve for the number of moles in the exercise:

• Convert pressure from mm Hg to atm. Use the conversion factor \(1 \, atm = 760 \, mm \, Hg\) to get \(748 \, mm \, Hg = 0.9842 \, atm\).
• Convert the volume of gas from mL to L for easier calculations, getting \(30.4 \, mL = 0.03 \, L\).
• Temperature in Celsius needs conversion to Kelvin by adding 273.15, resulting in \(296.15 \, K\)

By rearranging the equation to solve for \(n\), you can find the moles of gas that have reacted with the available ions to form a precipitate.

Solubility Product Constant

The solubility product constant (K_{sp}) is a vital concept in understanding the solubility of ionic compounds. It quantifies the equilibrium between the solid phase and its ions in solution. For a sparingly soluble compound, its \(K_{sp}\) can be expressed in terms of the molar concentrations of its constituent ions.

In the given problem, we deal with \(\mathrm{AgBrO}_{3}\), where the \(K_{sp}\) indicates how much this salt can dissolve:

• It is calculated by multiplying the equilibrium concentrations of the ions raised to the power of their coefficients in the balanced equation.
• For \(\mathrm{AgBrO}_{3}\), the equation is \(K_{sp} = [Ag^{+}][BrO_{3}^{-}]\).
• A small \(K_{sp}\) value, like \(0.00005041\), implies the compound is not very soluble.

Understanding \(K_{sp}\) helps predict whether a precipitate will form under given conditions.

Molar Concentration

Molar concentration, or molarity, measures the concentration of a solute in a given volume of solution. It is expressed in moles per liter (M or mol/L). When you know the number of moles and the volume of solution, you can easily calculate molarity.

In this particular exercise, we explore the molarity of \(\mathrm{Ag}^{+}\) and \(\mathrm{BrO}_{3}^{-}\) ions in the initial \(\mathrm{AgBrO}_{3}\) solution. This helps determine how much of the compound is present in a given solution.

• Determine moles through the stoichiometric reaction where \(\mathrm{H}_{2}\mathrm{S}\) reacted to precipitate \(\mathrm{Ag}^{+}\) ions.
• Convert the volume from mL to L, as in this case, the volume of \(\mathrm{AgBrO}_{3}\) was \(338 \, mL = 0.338 \, L\).
• Use the formula: \(\text{Molarity} = \frac{\text{moles}}{\text{volume in L}}\) to find concentration in mol/L.

This helps in calculating the \(K_{sp}\), giving insights into the compound's solubility tendency in water.