Question

A hypothetical weak acid, HA, was combined with \(\mathrm{NaOH}\) in the following proportions: \(0.20 \mathrm{~mol}\) of \(\mathrm{HA}\), \(0.080 \mathrm{~mol}\) of \(\mathrm{NaOH}\). The mixture was diluted to a total volume of \(1.0 \mathrm{~L}\), and the pH measured. (a) If \(\mathrm{pH}=4.80\), what is the \(\mathrm{p} K_{a}\) of the acid? (b) How many additional moles of \(\mathrm{NaOH}\) should be added to the solution to increase the \(\mathrm{pH}\) to \(5.00 ?\)

Short answer

(a) The pKa of the weak acid \(HA\) is approximately 4.48. (b) An additional 0.026 moles of NaOH are required to increase the pH of the solution to 5.00.

Step-by-step solution

Find the moles of HA and A- at equilibrium

The reaction between the weak acid, HA, and the strong base, NaOH, will look like this: \[HA + OH^- \rightarrow A^- + H_2O\] Initially, there are 0.20 moles of HA and 0.080 moles of NaOH. Since NaOH is a strong base, it will consume all the available hydroxide ions (OH⁻), producing an equal number of moles of the conjugate base, A⁻. Thus, the moles of each species after the reaction are: HA: \(0.20 - 0.080 = 0.12 \,mol\) A⁻: \(0.080 \, mol\)

Apply the Henderson-Hasselbalch equation for pKa

The Henderson-Hasselbalch equation relates the pH, pKa, and the ratio of the concentration of the conjugate base (A⁻) to the concentration of the weak acid (HA): \[pH = pKa + \log \frac{[A^-]}{[HA]}\] Given the pH of the solution as 4.80, we can now solve for the pKa value. First, compute the molar concentrations of A⁻ and HA by dividing the moles by the total volume of the solution (1.0 L): \([A^{-}] = \frac{0.080}{1.0} = 0.080\, M\) \([HA] = \frac{0.12}{1.0} = 0.12\, M\) Then, substitute the values into the Henderson-Hasselbalch equation: \[4.80 = pKa + \log \frac{0.080}{0.12}\] Solve for pKa: \[pKa = 4.80 - \log \frac{0.080}{0.12} \approx 4.48\]

Calculate how many moles of NaOH are needed to reach the desired pH

To find out how many moles of NaOH are needed to increase the pH of the solution to 5.00, we can use the Henderson-Hasselbalch equation again. This time, we know the pH and pKa, and need to find the new ratio of \(\frac{[A^-]}{[HA]}\): \[5.00 = 4.48 + \log \frac{[A^{-}]}{[HA]}\] First, solve the equation for the ratio: \[\frac{[A^{-}]}{[HA]} = 10^{5.00 - 4.48} \approx 10^{0.52} \approx 3.31\] Now, let's denote the additional moles of NaOH required as x. Since each mole of NaOH would generate one mole of A⁻ and consume one mole of HA, the new amounts of these species will be: A⁻: \(0.080 + x \,mol\) HA: \(0.12 - x \,mol\) Plug these values into the ratio in the Henderson-Hasselbalch equation: \[\frac{0.080+x}{0.12-x} = 3.31\] Now, solve the equation for x: \[x = \frac{0.080(3.31) - 0.12}{1 - 3.31} \approx 0.026\, mol\] So the answers are: (a) The pKa of the weak acid \(HA\) is approximately 4.48. (b) An additional 0.026 moles of NaOH are required to increase the pH of the solution to 5.00.

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a vital tool in understanding acid-base chemistry. It serves as a bridge between the pH of a solution and the pKa, which is the acid dissociation constant of a weak acid. The essence of the equation is to relate pH to the ratio of the concentrations of a weak acid (HA) and its conjugate base (A⁻).
The simple yet powerful equation is written as:
\[pH = pKa + \log \frac{[A^-]}{[HA]}\]
This formula indicates that by knowing two out of the three variables—pH, pKa, and the acid/base ratio—you can solve for the third. This is particularly helpful in buffer solutions where the concentration of acid and its conjugate base are comparable. In educational settings, this equation is used frequently to help students calculate the pH of buffer solutions upon addition of small amounts of acid or base. For example, if more base is added to the solution, the ratio \(\frac{[A^-]}{[HA]}\) increases, naturally resulting in a higher pH. Conversely, when more acid is added, this ratio decreases, leading to a lower pH. Understanding this equation allows students to predict and manipulate the pH of solutions in laboratory settings.

pKa calculation

The pKa value is an essential parameter in the realm of acid-base chemistry. It represents the acid dissociation constant (Ka) in logarithmic terms and is a measure of the strength of an acid in solution—the lower the pKa, the stronger the acid. Calculating pKa is critical when dealing with weak acids, as it helps predict how much the acid will dissociate and therefore, buffer a solution's pH against changes.
To actually calculate the pKa, you generally need to know the pH of the solution and the molar concentration of the weak acid and its conjugate base at equilibrium. Using the Henderson-Hasselbalch equation, you can express the pKa as follows:
\[pKa = pH - \log \frac{[A^-]}{[HA]}\]
In practical application, you may be given the pH of a solution and the initial amounts of a weak acid and its conjugate base, from which you can determine the pKa. This calculation is encapsulated in many textbook exercises and problems, such as the one provided, as it's essential for students to understand and be able to determine the strength of acids in various contexts.

Weak acid and strong base reaction

The interaction between a weak acid and a strong base is a fundamental concept in acid-base chemistry. When a weak acid (HA) reacts with a strong base, such as NaOH, the hydroxide ions (OH⁻) from the base will react completely with the acid to form water (H₂O) and the conjugate base (A⁻) of the weak acid:
\[HA + OH^- \rightarrow A^- + H_2O\]
This reaction creates a basic or alkaline solution due to the presence of the conjugate base. Since strong bases dissociate completely in solution, they shift the equilibrium of the reaction towards the products, thereby consuming the weak acid. The resulting solution will often be a buffer solution if there are significant amounts of both the weak acid and its conjugate base present. In the context of a classroom exercise, the initial moles of the weak acid and the strong base are given, allowing students to calculate the moles of each species at equilibrium. This information can then be used in conjunction with the Henderson-Hasselbalch equation to predict the pH of the resulting solution or to determine how additional amounts of the strong base will affect the solution's pH.