Question

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

Short answer

The pH of the given buffer solution is approximately 5.699. To make the pH equal to the pKa (6.0), we need to add NaOH. To achieve this, we would need to add 0.40 mol of NaOH to the 1.0 L buffer solution.

Step-by-step solution

Identify species in the buffer solution

In the buffer solution, we have a weak base, H2NNH2, and its conjugate acid, H2NNH3NO3. The concentrations are: [H2NNH2] = 0.40 M [H2NNH3NO3] = 0.80 M We would also need the pKa for H2NNH2. Searching a pKa table, we find that pKa = 6.0 for H2NNH2.

Calculate the pH of the buffer solution

We can use the Henderson-Hasselbalch equation to calculate the pH of the solution: \[pH = pKa + \log \frac{[A^-]}{[HA]}\] In this case, pKa = 6.0 [A-] = [H2NNH2] = 0.40 M (weak base) [HA] = [H2NNH3NO3] = 0.80 M (conjugate acid) Substitute these values into the equation: \[pH = 6.0 + \log \frac{0.40}{0.80}\]

Solve for pH

Calculating the logarithm and solving for pH: \[pH = 6.0 + \log (0.5)\] \[pH = 6.0 - 0.3010\] \[pH = 5.699\] The pH of the buffer solution is approximately 5.699.

Determine the reagent to add

In order to make the pH equal to the pKa (6.0), we need to decide whether to add HCl (an acid) or NaOH (a base). Since the current pH is lower than the pKa (5.699 < 6.0), we need to increase the pH, so we should add a base, NaOH.

Calculate the moles of NaOH required

We can use the equation: \[moles\;NaOH = ([HA] - [A-]) * volume\] Because we want the resulting pH to be equal to the pKa, the concentrations of the conjugate acid-base pair should be equal ([HA] = [A-]). So, we have: \[moles\;NaOH = (0.80 - 0.40) * 1.0 L\] \[moles\;NaOH = 0.40\;mol\] To achieve a pH equal to the pKa, you would add 0.40 mol of NaOH to the 1.0 L buffer solution.

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a mathematical relationship used in chemistry to estimate the pH of buffer solutions. A buffer solution is a mixture of a weak acid or a weak base and its conjugate salt that stabilizes the pH of a solution by neutralizing added acids or bases.

The equation is given by:
\[pH = pKa + \text{log} \frac{[A^-]}{[HA]}\]
Here, pKa is the acid dissociation constant, [A^-] is the concentration of the base form (the conjugate base), and [HA] is the concentration of the acid form (the conjugate acid) of the buffer system. This equation is particularly useful for calculating the pH when the concentrations of the conjugate acid and its corresponding base are known. Students can use this equation to understand how the pH will change when adjusting the ratio of [A^-] to [HA].

Conjugate Acid-Base Pair

A conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton (H+). In a buffer solution, we typically deal with a weak acid and its conjugate base, or a weak base and its conjugate acid.

The presence of both a weak acid and its conjugate base in a solution allows the buffer to resist changes in pH upon the addition of an acid or a base. For example, in the provided exercise, \(H_{2}NNH_{2}\) acts as a weak base, and \(H_{2}NNH_{3}NO_{3}\) is its conjugate acid. The mixture of these two substances with their respective concentrations maintains the stability of the solution's pH, which is vital for processes that require a constant pH environment.

pKa Value

The pKa value represents the negative base-10 logarithm of the acid dissociation constant (Ka) of a substance, which indicates how strongly an acid donates a proton to a base.

The formula to find pKa from Ka is:
\[pKa = -\text{log}_{10} (Ka)\]
In practical terms, the lower the pKa value, the stronger the acid because it implies a larger Ka, and hence a greater tendency to lose protons. In a buffer solution, knowing the pKa value is crucial to predicting the pH and to prepare solutions with desired pH levels by adjusting the ratio of the acid and its conjugate base. In our exercise, the pKa value of \(H_{2}NNH_{2}\) is provided as 6.0, which is essential for the calculations made using the Henderson-Hasselbalch equation.

Adding Reagents to Buffer

When working with buffer solutions, one might need to adjust the pH to a specific value for a particular application. This can be done by adding reagents, such as an acid (like HCl) or a base (like NaOH).

The decision on which reagent to add depends on whether the pH needs to be increased or decreased. If the pH is lower than the target, a base is added; if it's higher, an acid is required. The amount of reagent to add can be determined using stoichiometry and the principles of the buffer system.

In the context of our exercise, NaOH (a base) is added to increase the pH since the current pH (5.699) is below the target pKa (6.0). The amount of NaOH calculated reflects the moles needed to achieve the desired pH by equalizing the concentrations of the conjugate acid-base pair, thereby using the buffering capacity of the solution.