Problem 92 - Chemistry The Central Science

Question

The solubility of \(\mathrm{Cl}_{2}\) in \(100 \mathrm{~g}\) of water at \(\mathrm{STP}\) is \(310 \mathrm{~cm}^{3}\). Assume that this quantity of \(\mathrm{Cl}_{2}\) is dissolved and equilibrated as follows: $$ \mathrm{Cl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{Cl}^{-}(a q)+\mathrm{HClO}(a q)+\mathrm{H}^{+}(a q) $$ (a) If the equilibrium constant for this reaction is \(4.7 \times 10^{-4}\), calculate the equilibrium concentration of \(\mathrm{HClO}\) formed. \((\mathbf{b})\) What is the \(\mathrm{pH}\) of the final solution?

Step-by-Step Solution

Answer

The equilibrium concentration of HClO is approximately \(1.868 \times 10^{-2} \, moles\, L^{-1}\) and the pH of the final solution is approximately 1.73.

Step 1: Convert solubility of Cl2 to moles

Given that the solubility of Cl2 is 310 cm³ in 100 g of water (100 mL) at STP: 1 mol of any gas at STP = 22.4 L = 22400 cm³ Therefore, \( moles\, of\, Cl2 = \frac{310}{22400} = 0.013839 \, moles \)

Step 2: Set up the equilibrium expression and use ICE table

The balanced equation is: \( Cl2(aq) + H2O \rightleftharpoons Cl^{-}(aq) + HClO(aq) + H^{+}(aq) \) Let the concentration of Cl2 at equilibrium be x moles, then: - [Cl2] = \(0.013839 - x\) - [Cl-] = [HClO] = [H+] = x Equilibrium constant, K = \(4.7 \times 10^{-4}\) Now, setting up the equilibrium expression: \(K = \frac{[Cl^{-}][HClO][H^{+}]}{[Cl2]} \) Since [Cl-] = [HClO] = [H+], the expression becomes: \( K = \frac{x^3}{0.013839-x} \) Replacing K with the given value and solving for x: \( 4.7 \times 10^{-4} = \frac{x^3}{0.013839-x} \)

Step 3: Calculate equilibrium concentrations and pH

To find the equilibrium concentration of HClO and H+ ions, we need to solve the equation for x. In this case, we can use a numerical method or an approximation. For simplicity and practical purposes, we'll use an approximation because K is considerably small: Assuming \(x \ll 0.013839\), \( 4.7 \times 10^{-4} \approx \frac{x^3}{0.013839} \) Now, solving for x: \( x^3 = 4.7 \times 10^{-4} \times 0.013839 \) \( x = \sqrt[3]{6.503 \times 10^{-6}} \) \( x = 1.868 \times 10^{-2} \, moles\) Since [HClO] = [H+] = x, the equilibrium concentration of HClO is \(1.868 \times 10^{-2} \, moles\, L^{-1}\) and the concentration of H+ ions is also \(1.868 \times 10^{-2} \, moles\, L^{-1}\). To find the pH of the resulting solution, use the relationship: \( pH = - \log[H^+] \) Thus, \( pH = - \log(1.868 \times 10^{-2}) \) \( pH \approx 1.73 \) Therefore, the equilibrium concentration of HClO is approximately \(1.868 \times 10^{-2} \, moles\, L^{-1}\) and the pH of the final solution is approximately 1.73.

Key Concepts

Solubility

Solubility is a fundamental concept in chemistry that describes the ability of a substance, the solute, to dissolve in a solvent, forming a solution. In this exercise, we focus on the solubility of chlorine gas (\(\mathrm{Cl}_{2}\)) in water. At STP (Standard Temperature and Pressure), it's given that 310 cm³ of chlorine gas can dissolve in 100 grams of water.

To understand this:

• Solute: Chlorine gas, \(\mathrm{Cl}_{2}\).
• Solvent: Water, \(\mathrm{H}_2\mathrm{O}\).
• Solution: The mixture of \(\mathrm{Cl}_{2}\) dissolved in water.

The solubility can be expressed in terms of moles by converting the volume of gas to moles using the molar volume (22.4 L or 22400 cm³ at STP). This conversion helps us analyze how much of the gas participates in the equilibrium reaction described later in the problem.

Equilibrium Constant

The equilibrium constant (\(K\)) is a crucial figure in chemical equilibrium that quantifies the ratio of the concentrations of products to reactants at equilibrium. For the reaction:

\[\mathrm{Cl}_{2}(aq)+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{Cl}^{-}(aq)+\mathrm{HClO}(aq)+\mathrm{H}^{+}(aq)\]
The equilibrium constant expression is given by
\[K = \frac{[\mathrm{Cl}^{-}][\mathrm{HClO}][\mathrm{H}^{+}]}{[\mathrm{Cl}_{2}]}\]
This ratio tells us how far the reaction proceeds towards products before reaching equilibrium.

• In this case, given \(K = 4.7 \times 10^{-4}\), it shows that at equilibrium, the products are much less concentrated than the reactants.
• The small value of \(K\) indicates that the reaction does not proceed far in the forward direction, so most chlorine remains undissociated in water.

Such understanding assists in predicting the behavior of a chemical system under equilibrium conditions.

pH Calculation

The pH of a solution measures its acidity or basicity and is calculated from the concentration of hydrogen ions (\([\mathrm{H}^{+}]\)). In this problem, the formation of \(\mathrm{HClO}\) and \(\mathrm{H}^{+}\) ions from the solubility and reaction of \(\mathrm{Cl}_{2}\) in water allows us to determine the pH.

To find the pH:

• Calculate the equilibrium concentration of \([\mathrm{H}^{+}]\), which from the solution is \(1.868 \times 10^{-2}\, \mathrm{moles\, L^{-1}}\).
• Use the formula \( pH = - \log [\mathrm{H}^{+}] \).

Here, plugging in the values gives:
\[pH = - \log (1.868 \times 10^{-2}) \approx 1.73\]
This low pH reflects the acidic nature of the solution due to the presence of hydrogen ions. Understanding the pH calculation helps in gauging the impact of solubility and equilibrium processes on acidity, which is vital in various chemical applications.