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 \(30.0-\mathrm{mL}\) 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.

Short answer

Question: Calculate the pH of a 10.0% (by mass) solution of hypochlorous acid (HClO) before titration, halfway to the equivalence point, and at the equivalence point when titrated with KOH. Answer: Before titration, the pH of the HClO solution is 4.11. Halfway to the equivalence point, the pH is 7.52. At the equivalence point, the pH of the solution is 9.97.

Step-by-step solution

Find the concentration of HClO solution before titration

We have a \(10.0\%\) (by mass) solution, meaning there are 10 g of HClO in 100 g of the solution. The density of the solution is \(1.00 \mathrm{~g} / \mathrm{mL}\), so 100 g of the solution is equal to 100 mL. Now, calculate the moles of HClO in the 100 mL solution: Molar mass of HClO = 35.5 (Cl) + 1.0 (H) + 16.0 (O) = \(52.5\,\mathrm{g/mol}\) Moles of HClO \(= \frac{10.0\,\mathrm{g}}{52.5\,\mathrm{g/mol}} = 0.1905\,\mathrm{mol}\) Now, calculate the concentration of HClO: \([HClO] = \frac{0.1905\,\mathrm{mol}}{100\,\mathrm{mL}} = 1.91\,\mathrm{M}\)

Calculate the pH before titration

To find the pH before titration, we can use the Ka expression for HClO using the given Ka value: \(Ka = \frac{[H^{+}][ClO^{-}]}{[HClO]}\) For the 1.91M HClO solution, the initial concentration of ClO- is nearly 0. At equilibrium, [H+]=[ClO-], let's denote it as x. The Ka of HClO is \(3.0\times10^{-8}\). \(Ka = \frac{x^{2}}{1.91-x}\). Considering x to be very small compared to 1.91, we can rewrite the equation as: \(3.0\times10^{-8} = \frac{x^{2}}{1.91}\) Solve for x, which represents the concentration of H+ ions: \(x = 7.69 \times 10^{-5} \mathrm{M}\) Now, calculate pH using the formula: \(pH = -\log([H^{+}])\) \(pH = -\log(7.69\times10^{-5}) = 4.11\) Before titration, the pH of the HClO solution is 4.11.

Calculate the pH halfway to the equivalence point

Halfway to the equivalence point implies that half the moles of HClO have reacted with KOH, forming ClO- ions. Use the Henderson-Hasselbalch equation to calculate the pH half-way to the equivalence point: \(pH = pK_a + \log(\frac{[ClO^{-}]}{[HClO]})\) The pKa is equal to -log(Ka): \(pK_a = -\log(3.0\times10^{-8}) = 7.52\) Now, plug in the known values: \(pH = 7.52 + \log(\frac{0.5 \times 1.91}{(1 - 0.5) \times 1.91}) = 7.52\) Halfway to the equivalence point, the pH is 7.52.

Calculate the pH at the equivalence point

At the equivalence point, the amount of moles of HClO = moles of KOH. We are given a \(30.0\,\mathrm{mL}\) sample of the HClO solution, which has a concentration of \(1.91\,\mathrm{M}\). Moles of HClO in the sample = \(1.91\,\mathrm{M} \times 30.0\,\mathrm{mL} = 0.0573\,\mathrm{mol}\). As the moles of HClO = moles of KOH, determine the volume of \(0.419\,\mathrm{M}\) KOH needed: Volume of KOH \(= \frac{0.0573\,\mathrm{mol}}{0.419\,\mathrm{M}} = 136.8\,\mathrm{mL}\) Now calculate the ClO- ion concentration at the equivalence point: \([ClO^{-}] = \frac{0.0573\,\mathrm{mol}}{166.8\,\mathrm{mL}} = 0.343\,\mathrm{M}\) But ClO- is a weak base and undergoes hydrolysis to form OH- ions. Write Kb for ClO-: \(K_b = \frac{[OH^{-}][HClO]}{[Cl^{-}]}\) To calculate Kb, use the \(K_w = K_a\times K_b\) relationship, where \(K_w = 1 \times 10^{-14}\) at 25°C: \(K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{3.0\times10^{-8}} = 3.33 \times 10^{-7}\) In the hydrolysis reaction, \([OH^{-}] = [HClO] = x\): \(3.33\times 10^{-7} = \frac{x^2}{0.343 - x}\) Assume x to be very small compared to 0.343: \(3.33\times 10^{-7} = \frac{x^2}{0.343}\) Solve for x, which represents the concentration of OH- ions: \(x = 9.28 \times 10^{-5}\,\mathrm{M}\) Now, calculate pOH using the formula: \(pOH = -\log([OH^{-}])\) \(pOH = -\log(9.28\times10^{-5}) = 4.03\) And finally, find pH using the relationship between pH and pOH: \(pH = 14 - pOH = 14 - 4.03 = 9.97\) At the equivalence point, the pH of the solution is 9.97.

Key concepts

Understanding Acid-Base Titration

Acid-base titration is a fundamental technique in chemistry used to determine the concentration of an unknown acid or base solution by reacting it with a base or acid of known concentration. When an acid and a base react, they neutralize each other to form water and a salt. This process is monitored using a pH meter or an indicator that shows a color change at a certain pH level.

During titration, a titrant (known solution) is slowly added to the analyte (unknown solution) until the reaction reaches the equivalence point, at which the numbers of moles of hydrogen ions equal the number of moles of hydroxide ions. pH calculations during different stages of titration help us to understand the progress of the acid-base reaction and the properties of the involved substances.

The Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a critical tool for pH calculation in buffer solutions and during titrations. It's given by: \[ pH = pK_a + \text{log}(\frac{[A^-]}{[HA]}) \]
where \(pK_a\) is the negative logarithm of the acid dissociation constant (\(K_a\)), \( [A^-] \) is the molar concentration of the conjugate base, and \( [HA] \) is the concentration of the acid. The equation is particularly useful halfway to the equivalence point of a titration because the concentrations of the acid and its conjugate base are equal, simplifying the equation to \( pH = pK_a \), since log(1) is 0. This allows for a straightforward calculation of the solution's pH at this particular stage.

Molarity: A Measure of Concentration

Molarity (\(M\)) is a unit of concentration that refers to the number of moles of a solute per liter of solution. It is one of the essential aspects of titration as it allows precise determination of how much titrant should be added to the analyte to reach the equivalence point. The calculation of molarity involves both the mass of the solute and the volume of the solution, as shown by: \[ [Solute] = \frac{moles\text{ of solute}}{volume\text{ of solution in liters}} \]
The molarity of the titrant directly influences the volume required to neutralize the analyte, which is crucial for reaching the equivalence point accurately.

Equivalence Point: The Target of Titration

The equivalence point in a titration occurs when the quantity of titrant added is stoichiometrically equivalent to the amount of substance in the analyte. This point is not necessarily when the pH of the solution is 7; it rather depends on the strength of the acids and bases involved. In the case of a strong acid-strong base titration, the equivalence point pH will be neutral, around 7. However, if a weak acid or base is involved, the pH will be above or below 7 respectively at the equivalence point.

To calculate the equivalence point during a titration, one must understand the reaction stoichiometry and use molarity calculations to determine the volume at which the reactants are perfectly neutralized. This point is often indicated by a sharp change in pH, which can be identified on a titration curve or by a color indicator that changes at or near the desired pH level.