Problem 135 - General Chemistry

Question

A 30.0 -mL sample of \(0.05 M \mathrm{HClO}\) is titrated by a \(0.0250 \mathrm{M} \mathrm{KOH}\) solution. \(K_{a}\) for \(\mathrm{HClO}\) is \(3.5 \times 10^{-8}\). Calculate a the \(\mathrm{pH}\) when no base has been added; \(b\) the \(\mathrm{pH}\) when \(30.00 \mathrm{~mL}\) of the base has been added; \(\bar{c}\) the \(\mathrm{pH}\) at the equivalence point; \(\square\) the \(\mathrm{pH}\) when an additional \(4.00 \mathrm{~mL}\) of the KOH solution has been added beyond the equivalence point.

Step-by-Step Solution

Answer

a) pH ≈ 3.88; b) pH ≈ 7.46; c) pH ≈ 9.93; d) pH ≈ 11.19.

Step 1: Find the initial pH of the HClO solution

To calculate the initial pH, use the equation \( K_a = \frac{[H^+][ClO^-]}{[HClO]} \). Since no base has been added yet, \([ClO^-] = [H^+]\), so:\[ K_a = \frac{[H^+]^2}{0.05 - [H^+]} \approx \frac{[H^+]^2}{0.05} \]Now, solve for \([H^+]\):\[ [H^+] = \sqrt{K_a \times 0.05} = \sqrt{3.5 \times 10^{-8} \times 0.05} \approx 1.32 \times 10^{-4} \]Finally, the pH is:\[ \text{pH} = -\log_{10}([H^+]) \approx -\log_{10}(1.32 \times 10^{-4}) \approx 3.88 \]

Step 2: Determine pH after adding 30.00 mL base

Determine the moles of HClO and KOH added. Moles of HClO = \(0.05 \times 0.030 = 0.0015 \) mol, and moles of KOH = \(0.0250 \times 0.030 = 0.00075 \) mol. After adding KOH, the remaining HClO is \(0.0015 - 0.00075 = 0.00075 \) mol. This leaves a buffer solution of HClO/ClO\(^-\). The pH can be found with the Henderson-Hasselbalch equation:\[ \text{pH} = \text{pK}_a + \log\left(\frac{[ClO^-]}{[HClO]}\right) \]The pK\(_a\) is:\[ \text{pK}_a = -\log_{10}(3.5 \times 10^{-8}) \approx 7.46 \]Therefore,\[ \text{pH} = 7.46 + \log\left(\frac{0.00075}{0.00075}\right) = 7.46 \]

Step 3: Find the pH at the equivalence point

At the equivalence point, all HClO has been converted to ClO\(^-\). This salt hydrolyzes to produce OH\(^-\) ions. Use the relation \( K_w = K_a \times K_b \) to find \( K_b \): \[ K_b = \frac{10^{-14}}{3.5 \times 10^{-8}} \approx 2.857 \times 10^{-7} \]The concentration of ClO\(^-\) in the solution is \(\frac{0.0015}{0.030 + 0.030} \approx 0.025 \) M. Solve for OH\(^-\):\[ [OH^-] = \sqrt{K_b \times [ClO^-]} = \sqrt{2.857 \times 10^{-7} \times 0.025} \approx 8.48 \times 10^{-5} \]Calculate the pOH:\[ \text{pOH} = -\log_{10}([OH^-]) = -\log_{10}(8.48 \times 10^{-5}) \approx 4.07 \]The pH is:\[ \text{pH} = 14 - \text{pOH} = 14 - 4.07 = 9.93 \]

Step 4: Determine pH after adding 34.00 mL of base

With additional 4.00 mL, calculate total moles KOH: \( 0.0250 \times 0.034 = 0.00085 \) mol. The amount exceeding HClO neutralizes is \(0.00085 - 0.0015 = 0.0001 \) mol. Solution volume is now 64 mL or 0.064 L. Compute OH\(^-\):\[ [OH^-] = \frac{0.0001}{0.064} \approx 0.0015625 \]The pOH is:\[ \text{pOH} = -\log_{10}(0.0015625) \approx 2.81 \]The corresponding pH is:\[ \text{pH} = 14 - 2.81 \approx 11.19 \]

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch Equation is a powerful tool for calculating the pH of buffer solutions. It is derived from the acid dissociation constant (Ka) in the context of a solution containing both acid and its conjugate base. The equation is:

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

For buffers, which resist changes in pH, the Henderson-Hasselbalch equation simplifies the computation of pH by considering the ratio of concentrations of the conjugate base to the weak acid. In our exercise, the solution after adding 30 mL of base is a buffer system of hypochlorous acid (\[ \text{HClO} \]) and its conjugate base, hypochlorite (\[ \text{ClO}^- \]).
By applying the Henderson-Hasselbalch equation, the pH is calculated as 7.46, given that the concentrations of the acid and base are equal. The significance lies in the ability of the buffer solution to maintain this pH, resisting drastic changes even with the addition of small amounts of acid or base.

pH Calculations

Calculating the pH of a solution is essential in understanding the acidity or basicity of that solution. The pH is the negative logarithm of the hydrogen ion concentration (\[ [H^+] \]), expressed as:

• \[\text{pH} = -\log_{10}([H^+])\]

In the exercise, we are handling different pH calculations for various stages of a titration. Starting with the pH of a pure \[ \text{HClO} \] solution, we calculated its initial pH by using its known Ka value and solving for \[ [H^+] \].
As the titration progresses, we adjusted pH calculations based on the changing nature of the solution, whether it is a buffer system or at the equivalence point. Advanced tools like the Henderson-Hasselbalch equation come into play during these calculations to simplify the process and yield accurate results.
Each stage of the titration involves specific pH calculations, involving reactions and concentrations of hydrogen ions at that precise moment, reflecting the underlying chemistry of acid-base reactions.

Equivalence Point

The equivalence point in a titration is a crucial stage where the amount of titrant added is stoichiometrically equivalent to the amount of analyte in the original solution. For acid-base titrations, this means that all of the acid has reacted with the base, or vice versa. At this point, the reaction has reached its endpoint, and additional titrant will cause significant changes in pH.
In the case of our exercise, the equivalence point is reached when enough \( \text{KOH} \) has been added to completely neutralize the \( \text{HClO} \) present. The pH at the equivalence point often depends on the nature of the salt formed; here, it is governed by the hydrolysis of \( \text{ClO}^- \) which slightly increases the pH into the basic range.
Instead of simple neutral water (pH 7), we find that the equivalence point for \( \text{HClO} \) and \( \text{KOH} \) results in a pH of approximately 9.93 due to the formation of a basic salt. Understanding this helps explain why pH jumps significantly after the equivalence point.

Buffer Solution

A buffer solution is one that can resist changes in pH upon the addition of small amounts of acid or base. This property is essential in many biological systems and industrial processes where maintaining a stable pH is crucial. Buffers are typically composed of a weak acid and its conjugate base or a weak base and its conjugate acid.
In our exercise, the system temporarily forms a buffer when equal molar concentrations of \( \text{HClO} \) and \( \text{ClO}^- \) are present after titration with \( \text{KOH} \). This equilibrium between the weak acid and its conjugate base stabilizes the pH at 7.46, according to the Henderson-Hasselbalch equation.
The power of a buffer system is in its ability to prevent significant pH changes, ensuring environments remain conducive to particular chemical reactions or processes. While buffers can't neutralize large amounts of strong acids or bases, they are effective at managing small perturbations.