Question

A buffered solution is made by adding \(50.0 \mathrm{~g} \mathrm{NH}_{4} \mathrm{Cl}\) to \(1.00 \mathrm{~L}\) of a \(0.75 M\) solution of \(\mathrm{NH}_{3}\). Calculate the \(\mathrm{pH}\) of the final solution. (Assume no volume change.)

Short answer

The pH of the final solution is 9.14.

Step-by-step solution

Calculate the moles of NH4Cl and NH3 in the solution

First, we'll find the moles of NH4Cl that are added: Moles of NH4Cl = mass / molar mass = \( \frac{50.0\,\text{g}}{53.49\,\text{g/mol}} \) = 0.935 moles Now we find the moles of NH3 in the solution: Moles of NH3 = concentration × volume = (0.75 mol/L) × (1.00 L) = 0.75 moles

Calculate the concentrations of NH4+ and NH3 after mixing

Assuming no volume change, the volume remains 1.00 L. Then, \[ [\mathrm{NH_4}^+] = \frac{0.935\,\text{moles}}{1.00\,\text{L}} = 0.935\,\text{M} \] \[ [\mathrm{NH_3}] = \frac{0.75\,\text{moles}}{1.00\,\text{L}} = 0.75\,\text{M} \]

Calculate the pKa for NH4+ using the pKb of NH3

First, we need to find the pKa of NH4+ using the formula: \[ pK_a + pK_b = 14 \] The pKb of NH3 is given as 4.75, so we can find pKa: \[ pK_a = 14 - pK_b = 14 - 4.75 = 9.25 \]

Calculate the final pH using the Henderson-Hasselbalch equation

Now we will use the Henderson-Hasselbalch equation to find the pH of the solution: \[ pH = pK_a + \log \frac{[\mathrm{conjugate\,base}]}{[\mathrm{weak\,acid}]} \] In our case, the conjugate base is NH3 and the weak acid is NH4+: \[ pH = 9.25 + \log \frac{0.75\,\text{M}}{0.935\,\text{M}} = 9.25 - 0.108 = 9.14 \] Therefore, the pH of the final solution is 9.14.

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a simple and powerful tool for calculating the pH of a buffer solution. This equation is especially helpful when you are dealing with a weak acid and its conjugate base, like in the example with ammonia \((\mathrm{NH}_3)\) and ammonium chloride \(\mathrm{NH}_4\mathrm{Cl}\). The equation states:\[pH = pK_a + \log \left( \frac{[\mathrm{conjugate\,base}]}{[\mathrm{weak\,acid}]} \right)\]Here, \(pK_a\) is a crucial value that measures the strength of the weak acid. The logarithmic term then adjusts the pH based on the ratio of concentrations of the conjugate base to the weak acid.

• \( [\mathrm{conjugate\ base}] \) is the concentration of the base form, \(\mathrm{NH}_3\) in this case.
• \( [\mathrm{weak\ acid}] \) is the concentration of the acid form, \(\mathrm{NH}_4^+\).

By using this equation, you can understand how changes in the ratio of these concentrations affect the pH of the solution.
The equation assumes certain simplifications, such as that ambient temperature is around \(25^\circ\, \text{C}\) and that activity coefficients are near one, which is a pretty fair estimate for many lab conditions.
Thus, with the given concentrations and using \(pKa = 9.25\), we calculated: \(pH = 9.14\).

pKa and pKb relationship

Understanding the relationship between \(pK_a\) and \(pK_b\) is critical when dealing with acid-base chemistry. \(pK_a\) corresponds to the dissociation of a weak acid, while \(pK_b\) is related to the dissociation of a weak base.\[\ pK_a + pK_b = 14\]This relationship is extremely useful for converting a given \(pK_b\) (for the base) into \(pK_a\) (for its conjugate acid), or vice versa.

• For example, the \(pK_b\) of ammonia \(\mathrm{NH}_3\) is 4.75. Hence, the \(pK_a\) of ammonium \(\mathrm{NH}_4^+\) is calculated using \(14 - 4.75 = 9.25\).
• This conversion allows us to use the Henderson-Hasselbalch equation for pH calculations of buffer solutions.

The \(pK_a\) and \(pK_b\) relationship is important because it highlights the complementarity in behavior between acidic and basic species in a conjugate pair.

molarity calculations

Molarity, denoted as \(M\), is a way of expressing the concentration of a solution. It is defined as the number of moles of a solute per liter of solution:\[M = \frac{\text{moles of solute}}{\text{liters of solution}}\]Accurate molarity calculations are fundamental in buffer solutions for determining the initial concentrations of the buffer components.
In this exercise, we calculated:

• For \(\mathrm{NH}_4\mathrm{Cl}\): Moles = \(\frac{50.0\, \text{g}}{53.49\, \text{g/mol}} = 0.935\, \text{moles}\)
• For \(\mathrm{NH}_3\): Moles = \(0.75\, \text{mol/L} \times 1.00\, \text{L} = 0.75\, \text{moles}\)

These moles are then used to determine the concentrations:

• Concentration of \(\mathrm{NH}_4^+\): \(0.935\, \text{moles/L} = 0.935\, \text{M}\).
• Concentration of \(\mathrm{NH}_3\): \(0.75\, \text{moles/L} = 0.75\, \text{M}\).

Understanding molarity is essential as it directly affects how the pH of a buffered solution can be calculated using the Henderson-Hasselbalch equation.

weak acid and conjugate base

A weak acid and its conjugate base pair form the backbone of a buffered solution. A weak acid is partially ionized in water, whereas its conjugate base is the result of this ionization.
In a buffer, the weak acid neutralizes added bases and the conjugate base neutralizes added acids, maintaining pH stability.

• \(\mathrm{NH}_4^+\) (ammonium ion) acts as the weak acid in our example.
• \(\mathrm{NH}_3\) (ammonia) is the conjugate base formed when \(\mathrm{NH}_4^+\) loses a hydrogen ion \((\mathrm{H}^+)\).

The solution resists drastic pH changes due to the equilibrium between the weak acid and its conjugate base. When a small amount of acid or base is added, the equilibrium shifts to counteract the change, maintaining overall pH balance.