Problem 51 - Chemistry: Principles and Reactions

Question

Consider a \(10.0 \%\) (by mass) solution of hypochlorous acid. Assume the density of the solution to be \(1.00 \mathrm{~g} / \mathrm{mL}\). A \(30.0\) -mL sample of the solution is titrated with \(0.419 \mathrm{M} \mathrm{KOH}\). Calculate the \(\mathrm{pH}\) of the solution (a) before titration. (b) halfway to the equivalence point. (c) at the equivalence point.

Step-by-Step Solution

Answer

Question: Calculate the pH of a 10.0% (by mass) hypochlorous acid (HOCl) solution with a density of 1.00 g/mL, at three different stages of titration with a KOH solution of concentration 0.419 M. Find the pH (a) before titration, (b) halfway to the equivalence point, and (c) at the equivalence point. Answer: The pH of the solution (a) before titration is approximately 4.12, (b) halfway to the equivalence point is 7.46, and (c) at the equivalence point is approximately 9.92.

Step 1: Calculate the number of moles of HOCl and OH- ions in the initial solution

We are given the mass percentage of hypochlorous acid (HOCl) in the solution as \(10.0\%\). Hypochlorous acid has a molar mass of \(52.46 \mathrm{g/mol}\). To get the number of moles of HOCl in the solution, we must first determine the mass of the solution. We are given that a \(30.0\, \mathrm{mL}\) sample is taken, and the density of the solution is \(1.00 \mathrm{g/mL}\). So, the mass of the solution is: \(30.0\, \mathrm{mL} \times 1.00\, \mathrm{g/mL} = 30.0\, \mathrm{g}\) Since \(10.0\%\) of the solution is HOCl, we have: \(\mathrm{mass\: of\: HOCl} = 10.0\%\, \times 30.0\, \mathrm{g} = 3.00\, \mathrm{g}\) Now, to find the number of moles of HOCl: \(\mathrm{moles\: of\: HOCl} = \frac{3.00\, \mathrm{g}}{52.46\, \mathrm{g/mol}} = 0.0572\, \mathrm{mol}\) Initially, there are no OH- ions in the solution, so the number of moles of OH- ions is \(0\).

Step 2: Calculate the concentration of HOCl in the initial solution

The initial solution has a volume of \(30.0\, \mathrm{mL} = 0.0300\, \mathrm{L}\). Thus, the concentration of HOCl in the initial solution is: \(\mathrm{[HOCl]} = \frac{0.0572\, \mathrm{mol}}{0.0300\, \mathrm{L}} = 1.91\, \mathrm{M}\)

Step 3: Calculate the dissociation constant \((K_a)\) of HOCl and use the ICE table to find the pH

Hypochlorous acid (HOCl) dissociates in water as: \(\mathrm{HOCl \rightleftharpoons H^+ + OCl^-}\) We know the dissociation constant \((K_a)\) for HOCl is \(3.5 \times 10^{-8}\). Using the ICE table for the reaction, let x be the concentration of H+ ions and OCl- ions at equilibrium: HOCl -> H+ + OCl- Initial: 1.91M 0 0 Change: -xM +xM +xM Equilibrium: 1.91M-x xM xM Setting up the equation for \(K_a\): \(K_a = \frac{[\mathrm{H^+}][\mathrm{OCl^-}]}{[\mathrm{HOCl}]} = \frac{x^2}{1.91-x}\) Solving for x, which is equal to the concentration of H+ ions, we get: \(x ≈ 7.52 \times 10^{-5}\, M\) Finally, calculate the pH of the solution before titration: \( \mathrm{pH} = - \log{[\mathrm{H^+}]} = -\log{(7.52 \times 10^{-5})} ≈ 4.12\) (b) Calculate the pH halfway to the equivalence point

Step 4: Calculate the number of moles of OH- ions halfway to the equivalence point and use the ICE table to find the pH

Halfway to the equivalence point means we have added half the number of moles of OH- ions required to react completely with HOCl. To find this number of moles, we divide the initial moles of HOCl by 2: Half the moles of OH- ions required = \(\frac{0.0572\, \mathrm{mol}}{2} = 0.0286\, \mathrm{mol}\) Now we have a \((\mathrm{HOCl/OCl^-})\) buffer system, and we can use the Henderson-Hasselbalch equation to calculate the pH: \(pH = pK_a + \log{\left(\frac{[\mathrm{OCl^-}]}{[\mathrm{HOCl}]}\right)}\) We know that: [K_a] for HOCl = \(3.5 \times 10^{-8}\), so \(pK_a = -\log{(3.5 \times 10^{-8})} ≈ 7.46\) Since we are halfway to the equivalence point, the ratio \([\mathrm{OCl^-}]/[\mathrm{HOCl}]\) is \(1\). Hence, the pH at the halfway point is equal to \(pK_a\): \(pH = 7.46\) (c) Calculate the pH at the equivalence point

Step 5: Calculate the number of moles of OH- ions at the equivalence point and use the ICE table to determine the resulting pH

At the equivalence point, the number of moles of OH- ions added is equal to the number of moles of HOCl: Moles of OH- ions added = \(0.0572\, \mathrm{mol}\) At the equivalence point, all of the HOCl has been converted to its conjugate base OCl-. We are given that the KOH solution has a concentration of \(0.419\, \mathrm{M}\). So, the volume of KOH solution required can be calculated by: \(V_{KOH} = \frac{0.0572\, \mathrm{mol}}{0.419\, \mathrm{M}} = 0.136\, \mathrm{L}\) The total volume of the solution at the equivalence point is the sum of the initial volume of the acid solution and the volume of the base (KOH) solution: \(0.0300\, \mathrm{L} + 0.136\, \mathrm{L} = 0.166\, \mathrm{L}\) Now, we calculate the concentration of OCl- in the solution at the equivalence point: \(\mathrm{[OCl^-]} = \frac{0.0572\, \mathrm{mol}}{0.166\, \mathrm{L}} = 0.344\, \mathrm{M}\) Since all of the HOCl has reacted, we can assume that all of the added OH- ions have also reacted, and we can ignore them in our calculations. Now, we can find the concentration of OH- ions produced due to the dissociation of OCl- and find the pOH using the expression for \(K_b\). We know that \(K_w = K_a \times K_b\), where \(K_w = 1.00 \times 10^{-14}\) is the ion product constant for water. We calculate \(K_b\) for OCl-: \(K_b = \frac{K_w}{K_a} = \frac{1.00 \times 10^{-14}}{3.5 \times 10^{-8}} = 2.86 \times 10^{-7}\) Solving for the concentration of OH- ions: \(\mathrm{[OH^-]} = \sqrt{K_b \times [\mathrm{OCl^-}]} = \sqrt{(2.86 \times 10^{-7})(0.344)} ≈ 8.37 \times 10^{-5}\, \mathrm{M}\) Finally, we can calculate the pOH and pH of the solution at the equivalence point: \(\mathrm{pOH} = - \log{[\mathrm{OH^-}]} = -\log{(8.37 \times 10^{-5})} ≈ 4.08\) \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 4.08 ≈ 9.92\) The pH of the solution (a) before titration is approximately \(4.12\) (b) halfway to the equivalence point is \(7.46\) (c) at the equivalence point is approximately \(9.92\).

Key Concepts

Hypochlorous Acid

Hypochlorous acid (HOCl) is a weak acid commonly used as a disinfectant in both household cleaning and water treatment applications. It is a member of the oxychlorine acids family and is known for its ability to kill pathogens effectively.

• HOCl is a weak acid, meaning it only partially dissociates in water into hydrogen ions (H⁺) and hypochlorite ions (OCl^-).
• The dissociation constant, also known as the acid dissociation constant (\(K_a\)), is a crucial value for hypochlorous acid since it reflects the strength of the acid. For HOCl, \(K_a\) is approximately \(3.5 \times 10^{-8}\), indicating it is indeed weak.
• In chemical equations, hypochlorous acid can be represented as HOCl ↔ H⁺ + OCl^-.

When titrating a solution containing hypochlorous acid with a strong base, such as potassium hydroxide (KOH), it's important to consider the properties of HOCl to predict the changes in pH during the reaction process.

Equivalence Point

The equivalence point in a titration is reached when the quantity of titrant added equals the quantity of substance initially present in the solution being titrated. It's essentially the point at which the acid has been completely neutralized by the base.

• In an acid-base titration like the one involving hypochlorous acid and KOH, the equivalence point can be determined using stoichiometry, which ensures that the moles of base added equals the moles of acid present in the solution.
• At the equivalence point, for a weak acid like HOCl, the solution comprises entirely of the conjugate base (OCl^-) produced after the acid is neutralized.
• The pH at the equivalence point will be greater than 7 because the OCl^- ion acts as a weak base, contributing to a basic environment.

Recognizing the equivalence point is crucial, as it informs us when to halt the titration. Beyond this point, any additional titrant will cause a significant jump in the pH of the solution.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a vital tool in chemistry that relates the pH of a solution to the pK_a and the ratio of the concentrations of an acid and its conjugate base.

• This equation is particularly useful for calculating the pH in buffer solutions, which consist of weak acids and their conjugate bases.
• The equation is expressed as: \(\mathrm{pH} = \mathrm{pK_a} + \log{\left(\frac{\left[\mathrm{A^-}\right]}{\left[\mathrm{HA}\right]}\right)}\), where \(\left[\mathrm{A^-}\right]\) is the concentration of the conjugate base, and \(\left[\mathrm{HA}\right]\) is the concentration of the acid.
• At the halfway to the equivalence point in a titration involving a weak acid like hypochlorous acid, the concentration of the acid equals the concentration of its conjugate base, making the ratio 1. This results in the pH being equal to the pK_a of the acid.

It is also helpful for understanding how the pH will change with the addition of small amounts of acid or base, a concept critical during titrations to anticipate the approach to the equivalence point.