Question

Calculate the \(\mathrm{pH}\) of the solution that results from each mixture. a. 50.0 \(\mathrm{mL}\) of 0.15 \(\mathrm{M} \mathrm{HCHO}_{2}\) with 75.0 \(\mathrm{mL}\) of 0.13 \(\mathrm{MNaCHO}_{2}\) b. 125.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{M} \mathrm{NH}_{3}\) with 250.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\)

Short answer

The \(pH\) of mixture a is approximately 3.85, and the \(pH\) of mixture b is approximately 9.55.

Step-by-step solution

Determine the Moles of Acid and Conjugate Base for Mixture a

First, calculate the moles of \(HCHO_2\) by multiplying its concentration by its volume converted to liters: \(0.15\ M \times 0.0500\ L = 0.0075\ moles\). Then, calculate the moles of \(NaCHO_2\) using its concentration and volume: \(0.13\ M \times 0.0750\ L = 0.00975\ moles\).

Determine the Mixture's Total Volume for Mixture a

Add the volumes of both solutions to find the total final volume of mixture a: \(50.0\ mL + 75.0\ mL = 125.0\ mL\), or \(0.1250\ L\) when converted to liters.

Use the Henderson-Hasselbalch Equation for Mixture a

Apply the Henderson-Hasselbalch equation to determine the \(pH\): \[pH = \(pK_a\) + \(\log\dfrac{[A^-]}{[HA]}\).\] The \(pK_a\) for \(HCHO_2\) is generally known or should be provided. Assuming \(pK_a = 3.74\), and using the moles of \(A^-\) (from \(NaCHO_2\)) and \(HA\) (from \(HCHO_2\)), \[pH = 3.74 + \(\log\dfrac{0.00975}{0.0075}\).\]

Calculate the \(pH\) of the Mixture for Mixture a

Calculate the \(pH\): \[pH = 3.74 + \(\log(1.3)\)\approx 3.74 + 0.11 = 3.85.\]

Determine the Moles of Base and Conjugate Acid for Mixture b

Calculate the moles of \(NH_3\) by multiplying the concentration by the volume in liters: \(0.10\ M \times 0.1250\ L = 0.0125\ moles\). Next, calculate the moles of \(NH_4Cl\) similarly: \(0.10\ M \times 0.2500\ L = 0.0250\ moles\).

Determine the Mixture's Total Volume for Mixture b

The total final volume of mixture b is the sum of the individual volumes: \(125.0\ mL + 250.0\ mL = 375.0\ mL\), or \(0.3750\ L\) when converted to liters.

Use the Henderson-Hasselbalch Equation for Mixture b

Using the Henderson-Hasselbalch equation for the \(NH_3/NH_4^+\) buffer system with \(pK_b\) of \(NH_3\) provided or known (for example, \(pK_b = 4.75\)), and converting to \(pK_a\) of \(NH_4^+\) using the relation \(pK_a = 14 - pK_b = 14 - 4.75 = 9.25\), we get: \[pH = 9.25 + \(\log\dfrac{[NH_4^+]}{[NH_3]}\).\] Since \(NH_4^+\) comes from \(NH_4Cl\), \[pH = 9.25 + \(\log\dfrac{0.0250}{0.0125}\).\]

Calculate the \(pH\) of the Mixture for Mixture b

Calculate the \(pH\): \[pH = 9.25 + \(\log(2)\)\approx 9.25 + 0.30 = 9.55.\]

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a valuable tool for calculating the pH of buffer solutions. It expresses the relationship between the pH of a solution and the pKa of the acid in its buffer system. The equation is as follows:
\[\begin{equation} pH = pK_a + log\frac{[A^-]}{[HA]} \end{equation}\]
In this expression, \(pH\) is the acidity or alkalinity of the solution, \(pK_a\) is the acid dissociation constant of the weak acid, \(\[A^-\]\) is the concentration of the conjugate base, and \(\[HA\]\) is the concentration of the weak acid. This equation basically tells us how the proportion of acid to its conjugate base will determine the solution's pH. For instance, when a buffer contains equal concentrations of an acid and its conjugate base, the pH equals the pKa of the acid.
Buffer systems are often made from a weak acid and its salt (conjugate base) or a weak base and its salt (conjugate acid). The exercise provided concerns itself with two buffer systems, \(HCHO_2/NaCHO_2\) and \(NH_3/NH_4Cl\), and uses the Henderson-Hasselbalch equation to calculate the pH after mixing solutions of acid and conjugate base, or base and conjugate acid.

Buffer Solutions

Buffer solutions play a critical role in maintaining pH levels within narrow limits. They are capable of resisting changes in pH upon the addition of small amounts of strong acid or base. This characteristic makes them extremely useful in chemical and biological systems where pH stability is essential.
To create a buffer solution, you need a weak acid or base along with its conjugate base or acid. It is this combination that provides the buffering action. When strong acid is added to a buffer, the weak base present in the buffer neutralizes it; conversely, when strong base is added, the weak acid neutralizes it. This neutralization process maintains a relatively stable pH.

Importance in Calculations

Buffer capacity, which is the measure of a buffer's resistance to pH change, is also a critical concept. It depends on both the absolute concentrations of the buffer components and their ratio. The exercise we are discussing exemplifies buffer action and requires an understanding of how component concentrations impact pH, especially when solutions are mixed together. By applying the Henderson-Hasselbalch equation, we can see how the moles of acid and conjugate base, as well as the total volume of the mixture, affect the resulting pH.

Acid-Base Equilibrium

Acid-base equilibrium refers to the balance between the concentrations of acids and bases in a solution. It's a dynamic balance between the forward and reverse reactions in which acids donate protons to bases.
At equilibrium, the rates of the forward and reverse reactions are equal. The equilibrium constant, Ka, quantitatively describes this balance for weak acids, and similarly, Kb for weak bases. In water, the dissociation of water itself sets up a particular equilibrium (Kw), serving as the basis for pH calculations. Provided exercises often require computations involving these equilibrium constants to calculate concentrations of individual species at equilibrium.

Application in pH Calculations

To fully grasp the concept within the context of the provided exercise, remember that each mixture combines a weak acid and its conjugate base or conversely, which allows us to define the buffer system at play. By understanding the acid-base equilibrium, particularly the dissociation constants \(Ka\) or \(Kb\), we can predict and calculate the resulting pH of mixtures. The relationship between \(Ka\), \(Kb\), and \(Kw\) is particularly relevant when working with the Henderson-Hasselbalch equation as it may necessitate converting \(Kb\) to \(Ka\), as the exercise demonstrates for \(NH_3/NH_4^+\) buffer system.
Through these concepts and practices, students will be better prepared to tackle real-world problems involving the dynamic and essential nature of acid-base chemistry.