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

The pKa of the weak acid HA is approximately \(2.82\) and to increase the pH to \(5.00\), an additional \(0.026\) moles of NaOH are required.

Step-by-step solution

Calculate the concentration of OH⁻ ions and convert it to H⁺ ions concentration

Given the amount of NaOH (\(0.080\) mol), we can calculate the concentration of OH⁻ ions: \[ OH^⁻ = \frac{0.080\,\text{mol}}{1.0\,\text{L}} = 0.080\,\text{M} \] Now, we are also given pH which is \(4.80\). We can obtain the concentration of H⁺ ions from the pH using the following formula: \[ [\text{H}^+]=10^{-\text{pH}} \Longrightarrow [\text{H}^+]=10^{-4.80} \]

Calculate the initial concentrations of HA and A⁻

Given the amount of HA (\(0.20\) mol), we can calculate the initial concentration of HA: \[ [\text{HA}] = \frac{0.20\,\text{mol}}{1.0\,\text{L}} = 0.20\,\text{M} \] Since \(\text{HA}\) and \(\text{NaOH}\) react in a \(1:1\) ratio, we can find the initial concentration of A⁻ using the amount of NaOH reacted: \[ [\text{A}⁻] = [\text{OH}⁻] = 0.080\,\text{M} \]

Calculate the Ka of the acid

Now, we can use the known concentration of H⁺ ions, the initial concentrations of HA, and A⁻ ions to determine the Ka of the acid: \[ \text{Ka}= \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \Longrightarrow \text{Ka} = \frac{10^{-4.80}\cdot 0.080}{0.20} \]

Determine the pKa of the acid

Now that we have the Ka, we can calculate the pKa using the following formula: \[ \text{pKa}=-\log{(\text{Ka})} \]

Calculate the additional concentration of NaOH needed to increase the pH

We know that the desired pH is \(5.00\). Using the known pKa value, we can initialize the Henderson-Hasselbalch equation to determine the required ratio of A⁻ to HA: \[ \mathrm{pH}=\mathrm{p} K_{a}+\log \frac{[\text{A}⁻]}{[\text{HA}]} \] Let \(x\) be the additional mol of NaOH. Therefore, we will have: \[ \frac{[\text{A}⁻]+x}{[\text{HA}]-x}=\frac{[\text{A}⁻]_{\text{new}}}{[\text{HA}]_{\text{new}}} \]

Convert the concentration of NaOH to moles

Now, we just need to solve the Henderson-Hasselbalch equation for \(x\) and calculate the additional number of moles of NaOH required to increase the pH to the desired level. This \(x\) value will give us the additional moles of NaOH needed.

Key concepts

pKa Calculation

Understanding the strength of an acid in solution is essential in chemistry, and the \textbf{pKa} value is one of the key indicators of an acid's strength. The pKa calculation is necessary when dealing with weak acids, as it allows us to determine the pH at which half of the acid is dissociated into its conjugate base.

The procedure to calculate pKa involves finding the acid dissociation constant (\textbf{Ka}) first. For a weak acid HA, which dissociates into H⁺ and A⁻ ions, Ka can be expressed using the equilibrium concentrations of these species: \[\begin{equation}Ka = \frac{[H^{+}][A^{-}]}{[HA]}\end{equation}\]In the given problem, after reacting with NaOH, we have the concentrations of H⁺ and A⁻ ions, allowing us to calculate the Ka. Once Ka is determined, the pKa is simply the negative logarithm of the Ka value: \[\begin{equation}pKa = -\log(Ka)\end{equation}\]Converting from Ka to pKa not only simplifies the numbers (as Ka values can be very small) but also aligns the scale with pH, making comparisons and calculations more intuitive. For example, a lower pKa value indicates a stronger acid as it shows a higher tendency to donate protons.

Henderson-Hasselbalch Equation

When you’re trying to understand how to control the pH of a solution, especially in buffers, the \textbf{Henderson-Hasselbalch equation} becomes a crucial tool. It relates the pH of a solution to the pKa of the acid and the ratio of the concentrations of the conjugate base (A⁻) and the weak acid (HA). The equation is given by: \[\begin{equation}pH = pKa + \log \left( \frac{[A^{-}]}{[HA]} \right)\end{equation}\]This equation implicitly assumes that the concentration of the acid and its conjugate base do not change significantly with added strong base or strong acid, which is a valid assumption for dilute solutions, such as the ones typically encountered in titration problems. The Henderson-Hasselbalch equation is particularly useful in predicting how the pH will change with the addition of acid or base. As part of the solution to our exercise, this equation helped estimate how much more NaOH was required to attain the desired pH increment. When solving such problems, remember that a logarithmic function's argument must be a ratio, not an absolute value, and that the ratio changes as the weak acid reacts with a strong base.

pH Calculation

The \textbf{pH calculation} is at the heart of understanding acidity and basicity in aqueous solutions. pH is a measure of the hydrogen ion concentration in a solution and is defined as: \[\begin{equation}pH = -\log([H^{+}])\end{equation}\]In the context of our problem, the pH tells us about the resulting acidity after a weak acid is partially neutralized by a strong base. Calculating pH from a given hydrogen ion concentration is straightforward, but the challenge often lies in determining those hydrogen ion concentrations in the first place. For weak acids, this means accounting for both the dissociation of the acid in water and the neutralization reaction with a base, like NaOH, in our example.

Once we have the pH, this value can inform further calculations, such as adjustments needed to reach a target pH or the creation of buffer solutions. For students, it's essential to become comfortable with logarithmic calculations and the relations between pH, pKa, and concentration ratios to navigate through these chemical concepts effectively.