Question

Use information from Appendix \(\mathrm{D}\) to calculate the \(\mathrm{pH}\) of (a) a solution that is \(0.150 \mathrm{M}\) in sodium formate \((\mathrm{HCOONa})\) and \(0.200 \mathrm{M}\) in formic acid \((\mathrm{HCOOH})\) (b) a solution that is \(0.210 \mathrm{M}\) in pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) and \(0.350 \mathrm{M}\) in pyridinium chloride \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHCl}\right) ;\) (c) a solution that is made by combining \(125 \mathrm{~mL}\) of \(0.050 \mathrm{M}\) hydrofluoric acid with \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) sodium fluoride.

Short answer

The pH of the three different solutions are as follows: (a) Sodium formate and formic acid solution has a pH of approximately 3.45. (b) Pyridine and pyridinium chloride solution has a pH of approximately 4.94. (c) Hydrofluoric acid and sodium fluoride solution has a pH of approximately 2.98.

Step-by-step solution

(a) Sodium Formate and Formic Acid Solution

To calculate the pH, we first need to find the pKa of formic acid. According to Appendix D, formic acid has a \(Ka = 1.8 \times 10^{-4}\). To find the pKa, take the negative logarithm of the Ka: \[pKa = -\log{Ka} = -\log{(1.8 \times 10^{-4})} \approx 3.74\] Now, apply the Henderson-Hasselbalch equation to determine the pH: \[pH = pKa + \log{\frac{[\text{HCOONa}]}{[\text{HCOOH}]}} = 3.74 + \log{\frac{0.150}{0.200}} \approx 3.45\] The pH of the sodium formate and formic acid solution is approximately 3.45.

(b) Pyridine and Pyridinium Chloride Solution

First, find the pKa of pyridine. According to Appendix D, pyridinium ion has a \(Ka = 5.9 \times 10^{-6}\). Take the negative logarithm of the Ka to find the pKa: \[pKa = -\log{Ka} = -\log{(5.9 \times 10^{-6})} \approx 5.23\] Using the Henderson-Hasselbalch equation, calculate the pH of the solution: \[pH = pKa + \log{\frac{[\text{C}_5\text{H}_5\text{N}]}{[\text{C}_5\text{H}_5\text{NHCl}]}} = 5.23 + \log{\frac{0.210}{0.350}} \approx 4.94\] The pH of the pyridine and pyridinium chloride solution is approximately 4.94.

(c) Hydrofluoric Acid and Sodium Fluoride Solution

Determine the molarity of HF and F- in the final solution: For Hydrofluoric Acid: \[[\text{HF}] = \frac{0.050 \ \text{M} \times 125 \ \text{mL}}{125 \ \text{mL} + 50 \ \text{mL}} = 0.0333 \ \text{M}\] For Sodium Fluoride: \[[\text{F}^-] = \frac{0.100 \ \text{M} \times 50 \ \text{mL}}{125 \ \text{mL} + 50 \ \text{mL}} = 0.0286 \ \text{M}\] Now, find the pKa of hydrofluoric acid. According to Appendix D, hydrofluoric acid has a \(Ka = 7.2 \times 10^{-4}\). Take the negative logarithm to find the pKa: \[pKa = -\log{Ka} = -\log{(7.2 \times 10^{-4})} \approx 3.14\] Using the Henderson-Hasselbalch equation, calculate the pH of the solution: \[pH = pKa + \log{\frac{[\text{F}^-]}{[\text{HF}]}} = 3.14 + \log{\frac{0.0286}{0.0333}} \approx 2.98\] The pH of the hydrofluoric acid and sodium fluoride solution is approximately 2.98.

Key concepts

pH calculation

The pH of a solution is a measure of the acidity or alkalinity of that solution. The pH scale ranges from 0 to 14, with 7 being neutral. Values below 7 indicate an acidic solution, while values above 7 suggest a basic (alkaline) solution. Calculating pH is especially important in chemical and biological processes, where it can influence reaction rates, solubility, and the function of enzymes and other biomolecules.

To find the pH, we can use the Henderson-Hasselbalch equation, which relates pH to the concentration of an acid and its conjugate base in a buffer solution. The formula is:\[ pH = pKa + \log{\frac{[\text{Base}]}{[\text{Acid}]}} \]

Here, \([\text{Base}]\) and \([\text{Acid}]\) represent the molar concentrations of the base and acid, respectively. The \(pKa\) is the negative logarithm of the acid dissociation constant \(Ka\) and represents the pH at which half of the acid is dissociated. This equation is particularly useful for calculating the pH of buffer solutions, where it easily handles the presence of both weak acids and their conjugate bases.

acid-base equilibrium

In chemistry, acid-base equilibrium refers to the state of balance in a solution where the rate of the forward reaction (acid donating protons) equals the rate of the reverse reaction (conjugate base accepting protons). This equilibrium is crucial in understanding how solutions buffer changes in pH.

For weak acids, dissociation in water is not complete, and equilibrium is established between the undissociated acid (HA) and the ions produced (H+ and A-). The dissociation constant, \(Ka\), quantifies the strength of the weak acid:\[ Ka = \frac{[H^+][A^-]}{[HA]} \]

When a weak acid and its conjugate base are present in a solution, they form a buffer, capable of maintaining a relatively stable pH by neutralizing added acids or bases. This is where the Henderson-Hasselbalch equation comes into play, allowing us to compute pH changes in such buffered systems, maintaining equilibrium between the acid and base forms.

buffer solutions

Buffer solutions are special mixtures designed to maintain a stable pH when small amounts of acids or bases are added. They consist of a weak acid and its conjugate base, or a weak base and its conjugate acid. The unique property of buffers is their ability to resist changes in pH, which is vital in many biological and chemical processes.

For example, in biological systems, buffers help keep the pH of blood stable around 7.4, a level necessary for physiological functions. They work through the principle of acid-base equilibrium: when an external acid is added, the conjugate base in the buffer will neutralize it; conversely, when a base is added, the weak acid will counterbalance it.

Buffer capacity is determined by the concentrations of the acid and its conjugate base. High concentrations provide more buffering action. The ability of buffer solutions to maintain a constant pH is elegantly captured by the Henderson-Hasselbalch equation, which calculates how the balance between the acid and base in the buffer shifts in response to changes, thus stabilizing the pH.