Problem 49 - Chemistry An Atoms First Approach

Question

Calculate the \(\mathrm{pH}\) of a solution that is \(0.40 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) and \(0.80 M \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3} .\) In order for this buffer to have \(\mathrm{pH}=\) \(\mathrm{p} K_{\mathrm{a}},\) would you add HCl or NaOH? What quantity (moles) of which reagent would you add to \(1.0 \mathrm{L}\) of the original buffer so that the resulting solution has \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}} ?\)

Step-by-Step Solution

Answer

The pH of the given buffer solution is approximately 7.10. To change the pH equal to the pKa value of 7.60, you need to add 0.40 moles of NaOH to 1.0 L of the original buffer solution.

Step 1: Find the Ka and pKa of the weak acid

First, we'll need to find the Ka (acid dissociation constant) value for the amine cation, \(\mathrm{H}_{2}\mathrm{NNH}_{3}^+\). Recall that Ka is related to Kb by the equation Ka = Kw/Kb, where Kw is the ionization constant of water (\(1.0 \times 10^{-14}\)). In this case, we are given that \(\mathrm{H}_{2}\mathrm{NNH}_{2}\) is a \(0.40 M\) weak base, and, as an amine, it has a Kb value of \(4.0 \times 10^{-7}\). To find Ka, we'll use the relationship mentioned above: Ka = Kw/Kb = \(\frac{1.0 \times 10^{-14}}{4.0 \times 10^{-7}}\) Now, we calculate the Ka and pKa values: \[Ka = \frac{1.0 \times 10^{-14}}{4.0 \times 10^{-7}} = 2.5 \times 10^{-8}\] Taking the negative logarithm of Ka gives us the pKa value: \[pKa = -\log(2.5 \times 10^{-8}) \approx 7.60\]

Step 2: Apply the Henderson-Hasselbalch equation

Now that we have the pKa value, we can apply the Henderson-Hasselbalch equation to calculate the pH of the buffer solution: \[pH = pKa + \log \left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\right)\] Here, \(\mathrm{A^-}\) represents the concentration of \(\mathrm{H}_{2}\mathrm{NNH}_{2}\) (the weak base), and \(\mathrm{HA}\) represents the concentration of \(\mathrm{H}_{2}\mathrm{NNH}_{3}^+\) (the weak acid) which can be approximated to be equal to the concentration of \(\mathrm{H}_{2}\mathrm{NNH}_{3}\mathrm{NO}_{3}\). The given concentrations are as follows: \([A^-] = 0.40 M \) \([HA] = 0.80 M \) Substituting the values into the equation: \[pH = 7.60 + \log \left(\frac{0.40}{0.80}\right)\] Calculating the pH: \[pH = 7.60 + \log(0.5) \approx 7.10\]

Step 3: Determine the reagent and moles to adjust pH

Since the pH of our buffer is 7.10, and to make the pH equal to the pKa (7.60), we need to increase the pH. To do this, we would need to add NaOH (a strong base). Now let's find the moles of NaOH required to adjust the pH. Starting with the Henderson-Hasselbalch equation and rearranging for \(\frac{[A^-]}{[HA]}\): \[\frac{[\mathrm{A}^-]}{[\mathrm{HA}]} = 10^{(pH-pKa)}\] We know the pH (\(7.60\)) and the pKa (\(7.60\)) values, so we can solve for the ratio: \[\frac{[\mathrm{A}^-]}{[\mathrm{HA}]} = 10^{(7.60-7.60)} = 10^0 = 1\] This means the concentrations of \(\mathrm{H}_{2}\mathrm{NNH}_{2}\) and \(\mathrm{H}_{2}\mathrm{NNH}_{3}\mathrm{NO}_{3}\) should be equal, and since the solution volume is \(1.0 L\), the moles of each species should also be equal. Since we initially have \(0.40 \, mol \mathrm{H}_{2}\mathrm{NNH}_{2}\) and \(0.80 \, mol \mathrm{H}_{2}\mathrm{NNH}_{3}\mathrm{NO}_{3}\), we need to add \(0.80 - 0.40 = 0.40 \, mol\) of NaOH to the solution to achieve a 1:1 ratio. In summary, the pH of the given buffer solution is 7.10. To change the pH to the pKa value of 7.60, we need to add 0.40 moles of NaOH to 1.0 L of the buffer solution.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a simple formula used to estimate the pH of a buffer solution. This equation relates the pH to the concentration of the weak acid and its conjugate base in the solution. It is expressed as:

\[pH = pKa + \log \left(\frac{[A^-]}{[HA]}\right)\]

Here, \([A^-]\) represents the concentration of the base, and \([HA]\) represents the concentration of the acid. This equation is handy for calculating the pH when you know the concentrations of the acid and base in the buffer.

• Useful for: Calculating the pH of buffers.
• Requires: Concentration values of acid and base, and pKa of the acid.

In our exercise, this equation helps calculate the buffer pH using given concentrations and the pKa value.

pKa and Ka Calculation

The pKa and Ka values are essential in understanding acid and base strength. The Ka represents the acid dissociation constant, which measures how well an acid dissociates into its ions in water. The smaller the Ka, the weaker the acid. To find pKa, which is the negative logarithm of Ka, you use the formula:

\[pKa = -\log(Ka)\]

In the exercise, we were given the Kb of a base to find Ka using:

\[Ka = \frac{Kw}{Kb}\]

Where \(Kw\) is the ionization constant of water (\(1.0 \times 10^{-14}\)). By using this relationship, we found Ka and then calculated pKa:

• Essential for: Finding pH of buffers, understanding acid strength.
• Helps to: Relate Ka values to pH and buffer systems.

Weak Acid-Base Equilibrium

Weak acids and bases do not fully dissociate in water, creating an equilibrium between the undissociated molecules and the ions. Understanding this equilibrium is crucial for buffer solutions.

• Equilibrium constant: For weak acids, represented by Ka.
• Incomplete dissociation: Leads to a balance between ions and undissociated molecules.

This equilibrium allows the solution to resist changes in pH when small amounts of acids or bases are added. In our exercise, the equilibrium concept helps explain why the buffer resists large pH changes.

Acid-Base Buffer Systems

Buffer systems are solutions that minimize pH changes when acids or bases are added. They consist of a mixture of a weak acid and its conjugate base (or vice versa).

• Components: Weak acid and conjugate base or weak base and conjugate acid.
• Purpose: Maintain stable pH within a certain range.

In the exercise, the buffer was composed of \(\mathrm{H}_{2}\mathrm{NNH}_{3}^+\) (weak acid) and \(\mathrm{H}_{2}\mathrm{NNH}_{2}\) (conjugate base). Adjustments were suggested to bring the buffer to its target pH by adding a strong base, \(\text{NaOH}\). By understanding buffer systems, we can see how adding specific amounts of acids or bases helps achieve the desired pH stability.