Question

After solid \(\mathrm{SrCO}_{3}\) was equilibrated with a buffer at \(\mathrm{pH} 8.6\), the solution was found to have \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-4} \mathrm{M}\), what is \(K_{\mathrm{sp}}\) of \(\mathrm{SrCO}_{3} ?\left(K_{\mathrm{a} 2}\right.\) for \(\mathrm{H}_{2} \mathrm{CO}_{3}=5.0 \times 10^{-11}\), \(\log 2=0.3,5.1 \times 0.196=1.0)\) (a) \(4.0 \times 10^{-8}\) (b) \(8.0 \times 10^{-8}\) (c) \(8.0 \times 10^{-10}\) (d) \(3.38 \times 10^{-8}\)

Short answer

\(4.0 \times 10^{-8}\)

Step-by-step solution

Identify the Relevant Equations and Constants

The solubility product constant (\(K_{\text{sp}}\)) of \(\mathrm{SrCO}_{3}\) is given by \(K_{\text{sp}} = [\mathrm{Sr}^{2+}][\mathrm{CO}_{3}^{2-}].\) Since the solution is buffered at \(\mathrm{pH} 8.6\), we can determine the concentration of \(\mathrm{CO}_{3}^{2-}\) using the given \(K_{\mathrm{a2}}\) for \(\mathrm{H}_{2}\mathrm{CO}_{3}\) and the relationship between \(\mathrm{pH}\), \(\mathrm{pOH}\), and \(\mathrm{pK}_{\mathrm{a}}\).

Calculate the Concentration of \([\mathrm{CO}_{3}^{2-}]\)

Since \(\mathrm{pH} = 8.6\), \(\mathrm{pOH} = 14 - \mathrm{pH} = 5.4\). The \(K_{\mathrm{a2}}\) expression for \(\mathrm{H}_{2}\mathrm{CO}_{3}\) dissociation into \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{H}^{+}\) is \(K_{\mathrm{a2}} = \frac{[\mathrm{H}^{+}][\mathrm{CO}_{3}^{2-}]}{[\mathrm{HCO}_{3}^{-}]}\). Since \(\mathrm{pH} = 8.6\), \(\mathrm{pH} = -\log [H^{+}]\), thus \(\left[ H^{+} \right] = 10^{-8.6} \) M. From the given \(K_{\mathrm{a2}}\), we can find the ratio of \(\frac{[\mathrm{CO}_{3}^{2-}]}{[\mathrm{HCO}_{3}^{-}]}\) using \(K_{\mathrm{a2}}\) and \(\left[ H^{+} \right] \), and assume that \(\left[ HCO_{3}^{-} \right]\approx\left[OH^{-}\right]\) due to the buffer.

Assume \(\left[ \mathrm{HCO}_{3}^{-} \right] \approx \left[ \mathrm{OH}^{-} \right]\)

Given the pH of 8.6 and using the \(\mathrm{pOH}\), find the \(\left[ \mathrm{OH}^{-} \right]\) from the equation \(\left[ \mathrm{OH}^{-} \right] = 10^{-\mathrm{pOH}} = 10^{-5.4} \) M. Using the given approximation \(5.1 \times 0.196 = 1.0\), which corresponds to \(\left[ \mathrm{CO}_{3}^{2-} \right] / \left[ \mathrm{OH}^{-} \right]\), we can directly state that \(\left[ \mathrm{CO}_{3}^{2-} \right]\) is equal to \(5.1 \times \left[ \mathrm{OH}^{-} \right]\). Therefore, \(\left[ \mathrm{CO}_{3}^{2-} \right] = 5.1 \times 10^{-5.4}\).

Calculate the Solubility Product Constant \(K_{\text{sp}}\)

With the concentrations of \(\mathrm{Sr}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\), calculate \(K_{\text{sp}}\): \(K_{\text{sp}} = [\mathrm{Sr}^{2+}][\mathrm{CO}_{3}^{2-}]\) equals \(2.0 \times 10^{-4} \) M \(\times 5.1 \times 10^{-5.4} \) M which is \(4.0 \times 10^{-8}\) when rounded to one significant figure

Key concepts

Equilibrium Constant

The equilibrium constant is a number that expresses the ratio of the concentrations of products to reactants, each raised to the power of their coefficients as they appear in the balanced chemical equation, when a chemical reaction has reached equilibrium. It provides valuable information about the position of the equilibrium and hence the extent of a reaction. In the context of solubility, the equilibrium constant is referred to as the solubility product constant, represented by \(K_{\text{sp}}\).

For a sparingly soluble compound such as \(\text{SrCO}_{3}\), the \(K_{\text{sp}}\) expression is given by \(K_{\text{sp}} = [\text{Sr}^{2+}][\text{CO}_{3}^{2-}]\). It is essential to understand that this constant is specific to a particular compound at a given temperature and can be used to predict whether a precipitate will form in a solution. A low \(K_{\text{sp}}\) value indicates a less soluble compound. When solid \(\text{SrCO}_{3}\) equilibrates in a buffered solution, \(K_{\text{sp}}\) can be calculated using the concentration of \(\text{Sr}^{2+}\) ions, which must be compared with the concentration of \(\text{CO}_{3}^{2-}\) ions, calculated from pH and relevant equilibrium expressions.

Buffer Solution pH

A buffer solution is a system that resists changes in pH when small quantities of an acid or a base are added. It typically consists of a weak acid and its conjugate base or a weak base and its conjugate acid. The pH of a buffer depends on the concentration ratio of these two components and their respective acid dissociation constant, \(K_a\).

The Henderson-Hasselbalch equation is often used to calculate the pH of a buffer solution: \(\text{pH} = \text{p}K_a + \log(\frac{[\text{conjugate base}]}{[\text{weak acid}]})\). This equation shows the direct relationship between the pH of a solution, the \(K_a\) value of the weak acid, and the concentration ratio of the conjugate base to the weak acid. In our exercise, the buffering capacity was pivotal in determining the carbonate ion concentration, which was necessary for calculating the solubility product constant for \(\text{SrCO}_{3}\).

Understanding the pH of a buffer and its mechanisms is crucial for solving problems related to solubility and precipitation in different chemically buffered scenarios.

Concentration Calculation

Concentration calculation is fundamental for chemists as it allows predicting and controlling the outcomes of reactions. It involves determining the amount of substance present in a given volume of solution, usually expressed in molarity (M), which is moles per liter. In solubility problems, calculating the concentrations of ions is crucial to find the solubility product constant.

In our step-by-step solution, we used the \(K_{a2}\) value of \(\text{H}_2\text{CO}_{3}\) to determine the carbonate ion concentration at the given pH. This involved using the principles of chemical equilibrium and buffer solution behavior to assume the concentration of bicarbonate ions (\(\text{HCO}_3^-\)) was approximately equal to the hydroxide ions concentration (\(\text{OH}^-\)). With this assumption, along with the pH and \(K_{a2}\) values, we calculated the carbonate concentration needed for the \(K_{\text{sp}}\) equation. A clear understanding of these concentration calculations is essential in predicting whether a precipitate will form and in numerous other applications in chemistry.