Question

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

Short answer

After calculations, the initial pH of the given solution is approximately \(4.72\). After adding \(0.10\) moles of \(\mathrm{HCl}\), the pH drops to approximately \(4.44\). Finally, after adding \(0.20\) moles of \(\mathrm{NaOH}\) to the original solution, the pH increases to approximately \(5.16\).

Step-by-step solution

Write the acid dissociation equation and the expressions for \(K_a\) and \(pH\).

The acid dissociation equation for the weak acid, \(\mathrm{HOC}_6\mathrm{H}_5\), is: \[\mathrm{HOC}_6\mathrm{H}_5 + \mathrm{H_2O} \rightleftharpoons \mathrm{OC}_6\mathrm{H}_5^- + \mathrm{H_3O}^+\] Given the expression for the acid dissociation constant \(K_a\) is: \[K_a = \frac{[\mathrm{OC}_6\mathrm{H}_5^-][\mathrm{H}_3\mathrm{O}^+]}{[\mathrm{HOC}_6\mathrm{H}_5]}\] The expression for the pH of this solution can be written as: \[\mathrm{pH} = \mathrm{p}K_a + \log \frac{[\mathrm{OC}_{6}\mathrm{H}_{5}^{-}]}{[\mathrm{HOC}_{6}\mathrm{H}_{5}]}\]

Substitute the given values and calculate the pH.

Using the information provided: \([\mathrm{HOC}_6\mathrm{H}_5] = 0.85\,\text{M}\), \([\mathrm{OC}_6\mathrm{H}_5^-]=0.80\, \text{M}\), \(K_a = 1.6\times10^{-10}\). We can find the \(pK_a\) as follows: \[pK_a = -\log K_a = -\log (1.6 \times 10^{-10}) \] And then calculate the pH using the Henderson-Hasselbalch equation: \[\mathrm{pH} = (\mathrm{-\log (1.6 \times 10^{-10})})+ \log \frac{0.80 \,\text{M}}{0.85\, \text{M}}\] By calculating, the \(\mathrm{pH}\) of the solution is approximately \(4.72\). #b.Calculate the pH after 0.10 mole of HCl has been added to the original solution. Assume no volume change on addition of HCl.#

Determine the reaction between the solution and HCl and find new concentrations.

When \(\mathrm{HCl}\) is added to the solution, it reacts with the conjugate base \(\mathrm{OC}_6\mathrm{H}_5^-\). The reaction is as follows: \[\mathrm{OC}_6\mathrm{H}_5^- + \mathrm{HCl} \rightarrow \mathrm{HOC}_6\mathrm{H}_5 + \mathrm{Cl}^-\] Given that 0.10 mole of HCl is added, new concentrations are: \([\mathrm{OC}_6\mathrm{H}_5^-] = 0.80\, \text{M} - 0.10\, \text{mole/L} = 0.70\, \text{M}\) \([\mathrm{HOC}_6\mathrm{H}_5] = 0.85\, \text{M} + 0.10\, \text{mole/L} =0.95\, \text{M}\)

Calculate the new pH after HCl addition.

Using the new concentrations in the Henderson-Hasselbalch equation: \[\mathrm{pH} = (\mathrm{-\log (1.6 \times 10^{-10})})+ \log \frac{0.70 \,\text{M}}{0.95\, \text{M}}\] By calculating, the new \(\mathrm{pH}\) after adding \(\mathrm{HCl}\) is approximately \(4.44\). #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.#

Determine the reaction between the solution and NaOH and find new concentrations.

When \(\mathrm{NaOH}\) is added to the original solution, it reacts with the weak acid \(\mathrm{HOC}_6\mathrm{H}_5\). The reaction is as follows: \[\mathrm{HOC}_6\mathrm{H}_5 + \mathrm{NaOH} \rightarrow \mathrm{NaOC}_6\mathrm{H}_5 + \mathrm{H}_2\mathrm{O}\] Given that 0.20 mole of NaOH is added, new concentrations are: \([\mathrm{OC}_6\mathrm{H}_5^-] = 0.80\, \text{M} + 0.20\, \text{mole/L} = 1.00\, \text{M}\) \([\mathrm{HOC}_6\mathrm{H}_5] = 0.85\, \text{M} - 0.20\, \text{mole/L} =0.65\, \text{M}\)

Calculate the new pH after NaOH addition.

Using the new concentrations in the Henderson-Hasselbalch equation: \[\mathrm{pH} = (\mathrm{-\log (1.6 \times 10^{-10})})+ \log \frac{1.00 \,\text{M}}{0.65\, \text{M}}\] By calculating, the new \(\mathrm{pH}\) after adding \(\mathrm{NaOH}\) is approximately \(5.16\).

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a crucial tool in chemistry for calculating the pH of buffer solutions. These are solutions that resist changes in pH when small amounts of acids or bases are added. The equation provides a straightforward method to determine the pH of a weak acid and its conjugate base or a weak base and its conjugate acid.

The equation is expressed as:
\[\text{pH} = \text{p}K_a + \log \frac{[\text{A}^-]}{[\text{HA}]}\]
where:

• \(\text{pH}\) is the measure of acidity or basicity of the solution.
• \(\text{p}K_a\) is the negative logarithm of the acid dissociation constant, \(K_a\).
• \([\text{A}^-]\) is the concentration of the conjugate base.
• \([\text{HA}]\) is the concentration of the acid.

This equation highlights the pH dependent on the ratio of the concentrations of the conjugate base to the acid. The equation assumes that both the acid and its conjugate base are present in appreciable amounts. Therefore, it’s very effective in calculating the pH of buffer solutions. Let's move on to understanding acid-base reactions.

acid-base reaction

Acid-base reactions are fundamental chemical interactions where an acid donates a proton (\(H^+\)), and a base accepts the proton. This interaction is key to understanding changes in pH, especially when buffers are involved.

In the provided exercise, we observe two separate acid-base reactions:

• When \(\text{HCl}\) (a strong acid) is added to the buffer: It reacts with the conjugate base \(\text{OC}_6\text{H}_5^-\), forming \(\text{HOC}_6\text{H}_5\) and \(\text{Cl}^-\).
• When \(\text{NaOH}\) (a strong base) is added: It reacts with the weak acid \(\text{HOC}_6\text{H}_5\), producing \(\text{NaOC}_6\text{H}_5\) and water \(\text{H}_2\text{O}\).

Each interaction slightly changes the concentrations of the acid and its conjugate base, and thus, affects the pH. The acid-base reactions in buffers usually balance each other out, preventing sharp changes in pH. This resistance is why buffers are systematically used in many chemical and biological applications.

pH calculation

Calculating the pH of a solution accurately involves a deep understanding of the Henderson-Hasselbalch equation, especially in buffer solutions. Let's break this down using the steps from the exercise:

• First, identify the \(K_a\) of the weak acid and convert it to \(\text{p}K_a\) using: \(\text{p}K_a = -\log K_a\).
• Apply the Henderson-Hasselbalch equation to calculate the initial pH before adding any acids or bases: \(\text{pH} = \text{p}K_a + \log \frac{[\text{OC}_6\text{H}_5^-]}{[\text{HOC}_6\text{H}_5]}\).

Adjusting concentrations after adding an acid (or a base) reflects a change in pH:

• When strong acids like \(\text{HCl}\) are added, they react and alter only the base concentration in the equation, lowering the pH.
• When strong bases like \(\text{NaOH}\) are added, they increase the base concentration, causing an increase in pH.

Using precise subtractions or additions to these concentrations and applying the equation again provides the new pH of the solution, demonstrating the buffer's ability to maintain pH stability effectively.