Question

Consider 1.0 \(\mathrm{L}\) of a solution that is 0.85 \(\mathrm{M} \mathrm{HOC}_{6} \mathrm{H}_{5}\) and 0.80 \(\mathrm{M} \mathrm{NaOC}_{6} \mathrm{H}_{5} .\left(K_{\mathrm{a}} \text { for } \mathrm{HOC}_{6} \mathrm{H}_{5}=1.6 \times 10^{-10} .\right)\) a. Calculate the pH of this solution. b. Calculate the pH after 0.10 mole of \(\mathrm{HCl}\) has been added to the original solution. Assume no volume change on addition of HCl. c. Calculate the pH after 0.20 mole of NaOH has been added to the original buffer solution. Assume no volume change on addition of NaOH.

Short answer

The initial pH of the buffer solution is \(9.77\). After adding \(0.10\) mole of HCl, the pH becomes \(9.66\). Finally, after adding \(0.20\) mole of NaOH, the pH is \(10.02\).

Step-by-step solution

Identify the initial concentrations of the buffer solution

For this step, we already know that the solution is 0.85 M HOC₆H₅ (acid) and 0.80 M NaOC₆H₅ (base). The Kₐ for HOC₆H₅ is 1.6 × 10^(-10).

Calculate the initial pH of the buffer solution

To calculate the initial pH, we'll use the Henderson-Hasselbalch equation: pH = pKₐ + log ([A⁻]/[HA]) First, we need to find pKₐ. pKₐ = -log(Kₐ), so pKₐ = -log(1.6 × 10^(-10)) = 9.80. Now, we can find the pH: pH = 9.80 + log(0.80/0.85) = 9.80 + (-0.03) = \(9.77\).

Calculate the resulting concentrations after adding HCl

We have added 0.10 mole of HCl, which will react with NaOC₆H₅ to form HOC₆H₅. We can use the following stoichiometric relation: NaOC₆H₅ + HCl → NaCl + HOC₆H₅ Moles of base consumed = 0.10 mol New moles of HOC₆H₅ = 0.85 + 0.10 = 0.95 New moles of NaOC₆H₅ = 0.80 - 0.10 = 0.70 Now calculate the new concentrations from the new moles (assuming no volume change): [HOC₆H₅] = 0.95 mol/L [NaOC₆H₅] = 0.70 mol/L

Calculate the pH after adding HCl

Now, we can use the Henderson-Hasselbalch equation again for part b: pH = pKₐ + log ([A⁻]/[HA]) pH = 9.80 + log(0.70/0.95) = 9.80 - 0.14 = \(9.66\).

Calculate the resulting concentrations after adding NaOH

We have added 0.20 mole of NaOH, which will react with HOC₆H₅ to form NaOC₆H₅. HOC₆H₅ + NaOH → NaOC₆H₅ + H2O Moles of acid consumed = 0.20 mol New moles of HOC₆H₅ = 0.85 - 0.20 = 0.65 New moles of NaOC₆H₅ = 0.80 + 0.20 = 1.00 Now calculate the new concentrations from the new moles (assuming no volume change): [HOC₆H₅] = 0.65 mol/L [NaOC₆H₅] = 1.00 mol/L

Calculate the pH after adding NaOH

Now, we can use the Henderson-Hasselbalch equation again for part c: pH = pKₐ + log ([A⁻]/[HA]) pH = 9.80 + log(1.00/0.65) = 9.80 + 0.22 = \(10.02\). To summarize the results: a. The initial pH of the buffer solution is 9.77. b. The pH after adding 0.10 mole of HCl is 9.66. c. The pH after adding 0.20 mole of NaOH is 10.02.

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an essential tool in chemistry for understanding buffer solutions and their pH levels. This equation is particularly handy in biochemistry and medicine for preparing buffer solutions that maintain a stable pH within a certain range.

Here's the equation: \[ \text{pH} = \text{pK}_a + \log \left( \frac{[A^-]}{[HA]} \right) \]

• \([A^-]\) represents the concentration of the conjugate base.
• \([HA]\) represents the concentration of the weak acid.
• pKₐ is the negative logarithm of the dissociation constant (Kₐ) and represents the strength of the acid.

To use this equation effectively, you must know the concentrations of the weak acid and its conjugate base, along with the Kₐ of the acid.
These inputs help you accurately calculate the pH of the buffer solution. The equation assumes that the ionization of the weak acid and its conjugate base are in a dynamic equilibrium, providing a resistance to pH changes, which is the hallmark of buffer solutions.

pH Calculation

Calculating the pH of a buffer solution is a straightforward process once you understand the relevant components.
For the initial solution in our exercise, given that the solution has 0.85 M HOC₆H₅ (weak acid) and 0.80 M NaOC₆H₅ (base), the Henderson-Hasselbalch equation becomes your go-to tool.

### Initial pH CalculationTo find the initial pH:1. Calculate pKₐ using the Kₐ value provided: \[ \text{pK}_a = -\log(1.6 \times 10^{-10}) = 9.80 \]2. Use the equation: \[ \text{pH} = 9.80 + \log \left( \frac{0.80}{0.85} \right) \] Simplifying this gives a pH of 9.77.

As the buffer solution's concentrations change with the addition of an acid or base, repeat the process using the updated concentrations to find the new pH.
Remember, even small changes in the concentrations can lead to noticeable changes in the pH due to the logarithmic nature of the equation.

Acid-Base Reactions

Understanding acid-base reactions is crucial when working with buffer solutions. These reactions involve the transfer of protons (H⁺) between species, leading to changes in the composition of the solution.
In our exercise:

• When HCl (a strong acid) is added, it reacts with the conjugate base NaOC₆H₅, decreasing the base's concentration while increasing that of the acid HOC₆H₅.
• The reaction is: \( \text{NaOC}_6\text{H}_5 + \text{HCl} \to \text{NaCl} + \text{HOC}_6\text{H}_5 \)
• When NaOH (a strong base) is added, it reacts with the weak acid HOC₆H₅, decreasing the acid's concentration while increasing the base, NaOC₆H₅.
• The reaction is: \( \text{HOC}_6\text{H}_5 + \text{NaOH} \to \text{NaOC}_6\text{H}_5 + \text{H}_2\text{O} \)

These reactions illustrate the buffer's ability to resist drastic pH changes by transforming added strong acids or bases into their weaker counterparts, which dampens shifts in hydrogen ion concentration.

Chemical Equilibrium

In the context of buffer solutions, chemical equilibrium plays a vital role in maintaining consistent pH levels. A buffer solution consists of a weak acid and its conjugate base, which are in equilibrium.

### Dynamic Equilibrium This equilibrium allows the solution to maintain a stable pH: - When an acid is added, the equilibrium shifts towards the formation of more weak acid, using up the added H⁺ ions. - Conversely, when a base is added, the equilibrium favors the formation of more conjugate base, using up the hydroxide ions (OH⁻).
This balance is what gives buffers their ability to moderate pH changes.

### Le Châtelier's Principle This principle describes how a system at equilibrium responds to disturbances: - The system shifts in a direction that reduces the effect of the disturbance. For a buffer:

• Adding acid will shift the equilibrium to the left, reducing the increase in H⁺.
• Adding base will shift it to the right, neutralizing the added OH⁻.

The robust nature of buffer systems is a practical application of equilibrium concepts, ensuring that solutions can withstand additions of external acids or bases without significant changes in pH.