Question

A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\). The student then adds \(13.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{pH}\) of the resulting solution is \(4.7\). How is the value of \(\mathrm{p} K_{\mathrm{a}}\) for the unknown acid related to \(4.7 ?\)

Short answer

The value of \(\mathrm{p}K_a\) for the unknown weak acid is related to the pH of the resulting solution through the expression \(\mathrm{p}K_a = 4.672\). This means the weak acid has a lower dissociation constant than the final pH, making it a weaker acid than the final solution.

Step-by-step solution

Determine the amount of moles of the weak acid, strong base, and excess strong acid.

First, we need to find the moles of NaOH and HCl. We can use the given volumes and molarity to do this. Moles of NaOH: \(moles = Molarity \times Volume\) \(moles_{NaOH} = 0.100 \mathrm{M} \times 25.0 \mathrm{mL} = 0.00250 \mathrm{mol}\) Moles of HCl: \(moles_{HCl} = 0.100 \mathrm{M} \times 13.0 \mathrm{mL} = 0.00130 \mathrm{mol}\) Since the weak acid (HA) is titrated with the NaOH completely, the amount of moles HA is equal to the amount of moles of NaOH. \(moles_{HA} = moles_{NaOH} = 0.00250 \mathrm{mol}\) We also have an excess of strong acid, HCl. The amount of remaining HCl can be found by subtracting the moles of NaOH from the moles of HCl: \(moles_{HCl_{excess}} = moles_{HCl} - moles_{NaOH} = 0.00130 \mathrm{mol} - 0.00250 \mathrm{mol} = -0.00120 \mathrm{mol}\) Since we cannot have negative moles, we must have made an error, and it must be that moles of NaOH = moles of HA + moles of HCl. So: \(moles_{HA} = moles_{NaOH} - moles_{HCl} = 0.00250 \mathrm{mol} - 0.00130 \mathrm{mol} = 0.00120 \mathrm{mol}\) Now we have the correct moles of HA and excess HCl.

Calculate the concentration of H+ ions using the pH.

The pH of the resulting solution is given as 4.7. We can use this value to find the concentration of H+ ions: \[H^+ = 10^{-pH} = 10^{-4.7} = 1.995\times10^{-5} \mathrm{M}\]

Write the expression for the weak acid dissociation and find the \(\mathrm{p}K_a\).

The dissociation of the weak acid HA into H+ and A- ions can be written as: \( HA \rightleftharpoons H^+ + A^-\) We can write the expression for the equilibrium constant, \(K_a\): \[K_a = \frac{[H^+][A^-]}{[HA]}\] Since we have the data for the concentration of HCl, we can calculate the resulting concentration of A- ions and HA: Concentration of HA: \([HA] = \frac{moles_{HA}}{Total \: volume} = \frac{0.00120 \: \mathrm{mol}}{25.0 \: \mathrm{mL} + 13.0 \: \mathrm{mL}} = 3.2258 \times 10^{-5} \: \mathrm{M}\) Concentration of A- ions: \([A^-] = \frac{moles_{HCl_{excess}}}{Total \: volume} = \frac{0.00130 \: \mathrm{mol}}{25.0 \: \mathrm{mL} + 13.0 \: \mathrm{mL}} = \frac{0.00130 \: \mathrm{mol}}{38.0 \: \mathrm{mL}} = 3.4211 \times 10^{-5} \: \mathrm{M}\) Now we can substitute these values into the \(K_a\) expression: \[K_a = \frac{(1.995\times10^{-5} \: \mathrm{M})(3.4211\times10^{-5} \: \mathrm{M})}{(3.2258\times10^{-5} \: \mathrm{M})} = 2.128 \times 10^{-5}\] Finally, we can find the \(\mathrm{p}K_a\): \[\mathrm{p} K_a = -\log_{10}(K_a) = -\log_{10}(2.128 \times 10^{-5}) = 4.672\] To summarize, the value of \(\mathrm{p}K_a\) for this unknown weak acid is related to the pH of the resulting solution through the expression \( \mathrm{p}K_a = 4.672 \), which implies that the weak acid has a lower dissociation constant than the final pH, making it a weaker acid than the final solution.

Key concepts

pH Calculation

The calculation of pH is critical when working with acid-base equilibria. pH is a numeric scale used to specify the acidity or basicity of an aqueous solution. To calculate the pH from a known concentration of hydrogen ions \( ext{H}^+\), we use the formula: \[ \text{pH} = -\log_{10}([\text{H}^+]) \]
For example, if the \(\text{H}^+\) concentration is \(1.995\times10^{-5} \,M\), the pH is calculated by taking the negative logarithm of this concentration, which gives a pH of approximately 4.7. This relationship allows us to connect the concentration of \(\text{H}^+\) in a solution directly to its pH, providing a convenient way to understand the solution's acidity.

pKa Determination

Determining the \(\text{p}K_a\) of a weak acid is essential to understanding its strength and behavior in a solution. \(\text{p}K_a\) is the negative logarithm of the acid dissociation constant \(K_a\). The \(K_a\) represents how easily a weak acid donates protons in a solution, thus measuring its strength. A lower \(\text{p}K_a\) indicates a stronger acid.
To find \(\text{p}K_a\), we first calculate \(K_a\) using the expression: \[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \] Once \(K_a\) is known, we calculate \(\text{p}K_a\) as \( -\log_{10}(K_a)\).
In our example, with a \(\text{pH}\) of 4.7, using the concentration of ions obtained from the titration, we determine the \(K_a\) and subsequently the \(\text{p}K_a\) as 4.672, which indicates how feebly the weak acid dissociates.

Acid-Base Equilibria

Acid-base equilibria are crucial to understanding chemical reactions involving the transfer of protons between reactants. They involve the balance between acid and base species present in a solution. The equilibrium is determined by the dissociation constant \(K_a\) for an acid and \(K_b\) for a base. This tells us the proportion of the acid or base that dissociates in solution.
For weak acids, the equilibrium is typically described by the reaction: \[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \] Here, \(\text{HA}\) represents the weak acid that partially dissociates into its conjugate base \(\text{A}^-\) and hydrogen ions \(\text{H}^+\).
Understanding these equilibria helps us predict the behavior of acids and bases in various chemical contexts, especially during titrations where the concentrations of involved species change dynamically.

Phenolphthalein End Point

Phenolphthalein is a common acid-base indicator used in titrations. It helps determine the end point of a titration, which is when the amount of titrant is exactly enough to neutralize the analyte solution. Phenolphthalein changes color from colorless in acidic solutions to pink in basic solutions.
The color change occurs in the pH range of approximately 8.2 to 10.0, making it suitable for titrations of weak acids with strong bases, since it indicates when a slight excess of the base has been added.
In a titration involving a weak acid and a strong base, like \( ext{NaOH}\), the solution becomes pale pink at the end point, showing that the acid has been neatly neutralized. Using phenolphthalein allows for accurate determination of when the titration is complete.