Question

Calculate the \(\mathrm{pH}\) of each of the following solutions \(\left(K_{a}\right.\) and \(K_{b}\) values are given in Appendix \(\left.D\right):\) (a) \(0.095 M\) propionic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)\), (b) \(0.100 \mathrm{M}\) hydrogen chromate ion \(\left(\mathrm{HCrO}_{4}\right]\), (c) \(0.120 \mathrm{M}\) pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\).

Short answer

Using the given values and equilibrium expressions, we can calculate the [H+] for propionic acid and hydrogen chromate ion and [OH-] for pyridine as follows: (a) Propionic acid: [H+] = x (b) Hydrogen chromate ion: [H+] = x (c) Pyridine: [OH-] = x Next, we can use these concentrations to find the pH values: (a) Propionic acid: \(pH = -\log[H+]\) (b) Hydrogen chromate ion: \(pH = -\log[H+]\) (c) Pyridine: \(pH = 14 - pOH\) = 14 - \(-\log[OH^-]\) Substitute the calculated [H+] or [OH-] concentrations to find the pH values for each solution.

Step-by-step solution

Identify the given values and equilibrium expressions for each solution

For each solution, we need the equilibrium expressions and known values such as the concentration and Ka/Kb values. (a) Propionic acid (C2H5COOH): Equilibrium expression: C2H5COOH ⇌ C2H5COO- + H+ Concentration: 0.095 M Ka: Find the Ka value for propionic acid in Appendix D. (b) Hydrogen chromate ion (HCrO4): Equilibrium expression: HCrO4- ⇌ CrO4^2- + H+ Concentration: 0.100 M Ka: Find the Ka value for hydrogen chromate ion in Appendix D. (c) Pyridine (C5H5N): Equilibrium expression: C5H5N + H2O ⇌ C5H5NH+ + OH- Concentration: 0.120 M Kb: Find the Kb value for pyridine in Appendix D.

Calculate the [H+] or [OH-] concentration for each solution using the equilibrium expressions

For each solution, set up the equilibrium expression using the provided concentration and Ka/Kb values, and solve for the unknown [H+] or [OH-] concentration. (a) Propionic acid: \(K_a = \frac{[C2H5COO^-][H+]}{[C2H5COOH]}\) Assuming x moles of C2H5COOH ionize, we can write the equilibrium concentrations as follows: C2H5COOH: 0.095 - x C2H5COO-: x H+: x Plug in the values and solve for x ([H+]). (b) Hydrogen chromate ion: \(K_a = \frac{[CrO4^{2-} ][H+]}{[HCrO4^-]}\) Again, assume x moles of HCrO4- ionize: HCrO4-: 0.100 - x CrO4^2-: x H+: x Plug in the values and solve for x ([H+]). (c) Pyridine: \(K_b = \frac{[C5H5NH+][OH^-]}{[C5H5N]}\) Assume x moles of C5H5N react: C5H5N: 0.120 - x C5H5NH+: x OH-: x Plug in the values and solve for x ([OH-]).

Calculate the pH of each solution using the [H+] or [OH-] concentration

Use the calculated [H+] or [OH-] concentrations from Step 2 to find the pH of each solution. (a) Propionic acid: \(pH = -\log[H+]\) (b) Hydrogen chromate ion: \(pH = -\log[H+]\) (c) Pyridine: Since we found [OH-] for pyridine, first find the pOH using: \(pOH = -\log[OH^-]\) Then, find the pH using \(pH = 14 - pOH\) Finally, we have the pH values for each solution.

Key concepts

Acid Dissociation Constant (Ka)

Understanding the acid dissociation constant, represented by Ka, is crucial for pH calculation. The Ka value is a measure of the strength of an acid in solution. It is the equilibrium constant for the dissociation of a weak acid into its constituent ions. For example, a generic weak acid HA dissociates into H+ and A- ions, and its Ka expression is given by:
Ka = \(\frac{[H^+][A^-]}{[HA]}\).

In the pH calculation exercise, the Ka value of propionic acid and hydrogen chromate ion is needed to find the concentration of hydronium ions (H+). Generally, the lower the Ka value, the weaker the acid and less it dissociates in solution.

Base Dissociation Constant (Kb)

In contrast to an acid dissociation constant, the base dissociation constant (Kb) quantifies the strength of a base. It refers to the equilibrium constant for the dissociation of a weak base into its hydroxide ion (OH-) and its corresponding cation. The general equilibrium for a base B in water is:
B + H2O ⇌ BH+ + OH-
and the corresponding Kb is: Kb = \(\frac{[BH^+][OH^-]}{[B]}\).

For the given exercise, knowing the Kb value of pyridine is imperative to calculate the equilibrium concentration of hydroxide ions. From there, finding the pOH and eventually the pH is possible. The higher the Kb, the stronger the base and the more it will dissociate to release OH- ions in solution.

Equilibrium Concentration

Equilibrium concentration refers to the concentrations of reactants and products in a chemical reaction when the reaction has reached a state of dynamic equilibrium. At this point, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations remain constant over time. In pH calculations, we often assume that a weak acid or base only partially dissociates and use the initial concentration minus the change in concentration (usually denoted by 'x') to express the equilibrium concentrations of all species involved. This approach allows us to solve for 'x', which represents either [H+] or [OH-], depending on whether we're dealing with an acid or a base, respectively.

Propionic Acid

Propionic acid, or propanoic acid, is a weak organic acid with the chemical formula \(C_2H_5COOH\). It's important to note that weak acids do not fully dissociate in solution, making the calculation of pH less straightforward than for strong acids. For the exercise at hand, the concentration of propionic acid is given, and using its Ka value, its dissociation into propionate ions (\(C_2H_5COO^-\)) and hydronium ions (H+) can be quantified. Solving for the equilibrium concentration of H+ enables us to calculate the pH of the solution.

Hydrogen Chromate Ion

The hydrogen chromate ion (\(HCrO_4^-\)) is the conjugate base of chromic acid and a weak acid in its own right. Its equilibrium involves dissociation into the chromate ion (\(CrO_4^{2-}\)) and a hydronium ion (H+). To find the pH of a solution containing hydrogen chromate ion, we apply the acid dissociation constant (Ka) in a similar manner to other weak acids. Knowing this ion's concentration and Ka value, we can calculate the concentration of H+ ions at equilibrium, which is necessary to determine the pH of the solution.

Pyridine

Pyridine is a basic organic compound with a formula of \(C_5H_5N\). It acts as a weak base in aqueous solution, meaning it only partially accepts protons from water to form its conjugate acid (\(C_5H_5NH^+\)) and hydroxide ions (OH-). To find the pH of a pyridine solution, we first need to use the base dissociation constant (Kb) to calculate the equilibrium concentration of OH- ions. With this information, we can compute the pOH of the solution which, when subtracted from 14 (the sum of pH and pOH at 25°C), provides us the pH value.

pOH

The pOH of a solution is a measure of the hydroxide ion concentration and is very useful in calculations involving basic solutions. It is the negative base-10 logarithm of the molar concentration of hydroxide ions:
pOH = -log[OH-]

To find the pH from pOH, we use the following relationship at 25°C, where the sum of pH and pOH equals 14:
pH = 14 - pOH

This relationship allows us to convert pOH to pH, which is necessary when dealing with basic solutions like pyridine, as seen in the provided exercise.