Problem 144 - General Chemistry

Question

A \(0.150 \mathrm{M}\) solution of \(\mathrm{NaClO}\) is prepared by dis solving \(\mathrm{NaClO}\) in water. A \(50.0-\mathrm{mL}\) sample of this solution is titrated with \(0.100 \mathrm{M} \mathrm{HCl}\). Calculate the \(\mathrm{pH}\) of the solution at each of the following points of the titration: a prion to the addition of any \(\mathrm{HCl} ; \quad \emptyset\) halfway to the equivalence point; \(c\) at the equivalence point; di after \(5.00 \mathrm{~mL}\) of \(\mathrm{HC}\) has been added beyond the equivalence point. \(K_{a}\) for \(\mathrm{HClO}\) is \(3.5 \times 10^{-8}\)

Step-by-Step Solution

Answer

Initial pH: 9.82; Halfway pH: 7.46; Equivalence pH: 3.83; Beyond pH: 2.41.

Step 1: Calculate Initial pH

Before adding any HCl, the solution contains only NaClO. The \ClO\(^-\) ion acts as a base: \\[\text{ClO}^- + \text{H}_2\text{O} \rightleftharpoons \text{HClO} + \text{OH}^-\]First, calculate the concentration of \(\text{OH}^-\) using \(K_b\) for \(\text{ClO}^-\):\[K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-8}} \approx 2.86 \times 10^{-7}\]Use the equation for base ionization,\[K_b = \frac{[\text{OH}^-][\text{HClO}]}{[\text{ClO}^-]}\]Assume \([\text{HClO}] = [\text{OH}^-] = x\) and \([\text{ClO}^-] = 0.150 - x \approx 0.150 \ \text{M}\):\[2.86 \times 10^{-7} = \frac{x^2}{0.150}\]Solving for \(x\), we find \(x \approx 6.55 \times 10^{-5} \ \text{M}\). Hence \([\text{OH}^-] = 6.55 \times 10^{-5} \ \text{M}\).Finally, calculate the pH:\[pOH = -\log([\text{OH}^-]) \approx 4.18\]\[pH = 14 - pOH \approx 9.82\]

Step 2: Halfway to Equivalence Point

At the halfway point to equivalence, half of the ClO\(^-\) has been converted to its conjugate acid, HClO, resulting in a buffer solution. In buffers, \(\text{pH} = \text{pKa}\):\[pK_a = -\log(3.5 \times 10^{-8}) \approx 7.46\]Therefore, \(\text{pH} \approx 7.46\).

Step 3: Equivalence Point

At the equivalence point, all ClO\(^-\) is converted to HClO. We have an acidic solution due to the hydrolysis of HClO. Since the volume of titrant added equals 75.0 mL:Calculate \([\text{HClO}]\)\[[\text{HClO}] = \frac{0.075 \times 0.1}{0.050 + 0.075} = \frac{0.0075}{0.125} = 0.06 \ \text{M}\]Hydrolysis reaction of HClO:\[\text{HClO} + \text{H}_2\text{O} \rightleftharpoons \text{ClO}^- + \text{H}_3\text{O}^+\]Using \(K_a\), find \[\text{H}_3\text{O}^+\]:\[3.5 \times 10^{-8} = \frac{x^2}{0.06}\]Solving for \(x\), we find \(x \approx 1.48 \times 10^{-4} \ \text{M}\). Therefore, \([\text{H}_3\text{O}^+] = 1.48 \times 10^{-4} \ \text{M}\) and:\[pH = -\log(1.48 \times 10^{-4}) \approx 3.83\]

Step 4: Beyond Equivalence Point

Adding 5.00 mL beyond equivalence increases the volume to 130 mL. The excess HCl determines the pH:Calculate moles of excess HCl:\[Moles = 0.005 \times 0.1 = 0.0005 \ \text{mol}\]Concentration of excess \(\text{H}^+\):\[[\text{H}^+] = \frac{0.0005}{0.130} \approx 3.85 \times 10^{-3} \ \text{M}\]Calculate the pH:\[pH = -\log(3.85 \times 10^{-3}) \approx 2.41\]

Key Concepts

Buffer Solution

A buffer solution is a special kind of solution that resists changes in its pH when small amounts of an acid or a base are added. It is composed of a weak acid and its conjugate base, or a weak base and its conjugate acid. This unique property makes buffers very useful in many chemical and biological applications, as they can maintain a stable pH environment.

In the context of acid-base titration, a buffer solution can form halfway to the equivalence point. When titrating a weak base, like \( ext{ClO}^- ext{,}\) with a strong acid, like \( ext{HCl}\), the solution may form a buffer with equal concentrations of the weak base \( ext{ClO}^-\) and its conjugate acid \(\text{HClO}\). This situation typically occurs halfway through the titration process.

Buffer solutions are crucial in chemistry because they help predict titration outcomes and control the pH of various chemical environments. If you understand the buffer action, you can make accurate pH predictions and adjustments in experiments and industrial processes.

Equivalence Point

In an acid-base titration, the equivalence point marks the stage where the quantity of acid added equals the quantity of base present in the solution. At this point, the moles of acid are precisely equal to the moles of base, and the complete neutralization takes place.

For the titration of \(\text{NaClO}\) with \(\text{HCl}\), the equivalence point is reached when all \(\text{ClO}^-\) ions are converted into \(\text{HClO}\). During titration of a weak base with a strong acid, the solution will be slightly acidic at the equivalence point. This occurs because the weak base has been converted to its conjugate acid, thus lowering the pH. Calculating the equivalent point is crucial in titrating solutions, as it provides insights into the amount of titrant needed to entirely neutralize the analyte.

To determine the pH at the equivalence point, you need to understand the concentrations and possible hydrolysis reactions that occur afterward. For example, here the \(\text{HClO}\) can undergo hydrolysis, influencing which species are prominent in the solution after the equivalence point.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an essential formula in chemistry that helps you find the pH of a buffer solution. It is particularly handy in systems where the amount of acid-base pairs are known and need a quick estimation of the pH.

Here’s the equation:

• \[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]

This equation works best when the concentrations of the acid and conjugate base are relatively close to each other, as seen at the halfway point in titration, where \(\text{pH} = \text{pKa}\).

In practice, the Henderson-Hasselbalch equation is useful for understanding and predicting changes in pH during titrations and in buffer preparations. It shows how the equilibrium between acidic and basic components of a solution dictates its pH. Using this formula helps chemists and students accurately configure buffer solutions in labs and various applications.