Question

A solution is prepared by dissolving \(0.56 \mathrm{~g}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make \(1.0 \mathrm{~L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\), and the \(\mathrm{pH}\) of this solution.

Short answer

The concentrations and pH of the solution are as follows: \[C_6H_5CO_2H \approx 0.00458 \mathrm{~M}\] \[C_6H_5CO_2^- \approx 0.000585 \mathrm{~M}\] \[H^+ \approx 0.000585 \mathrm{~M}\] \[OH^- \approx 1.71 \times 10^{-11} \mathrm{~M}\] \[pH \approx 3.23\]

Step-by-step solution

Calculate the initial concentration of benzoic acid

First, we need to convert the mass of benzoic acid to moles and then divide by the volume of the solution to get the initial concentration of the benzoic acid. The molar mass of benzoic acid is about 122.12 g/mol. Moles of benzoic acid = \(\frac{0.56 ~g}{122.12 ~g/mol} = 0.00458 ~mol\) Initial concentration of benzoic acid = \(\frac{0.00458 ~mol}{1.0 ~L} = 0.00458 ~M\)

Write the dissociation reaction of benzoic acid and set up the equilibrium expression

The dissociation reaction of benzoic acid is given by: \(C_6H_5CO_2H \rightleftharpoons C_6H_5CO_2^- + H^+\) Let the change in concentration of benzoic acid be x. Then, the equilibrium concentrations will be: [benzoic acid] = \((0.00458 - x)~M\) [benzoate ion] = \(x~M\) [hydrogen ion] = \(x ~M\) The equilibrium expression with Ka value is: \(K_a = \frac{[C_6H_5CO_2^-][H^+]}{[C_6H_5CO_2H]}\)

Solve for concentrations and pH

Let's substitute the expressions for the equilibrium concentrations into the Ka expression: \(6.4 \times 10^{-5} = \frac{x^2}{0.00458 - x}\) As Ka is very small, we can make the assumption that \(x << 0.00458\), so the equation simplifies to: \(6.4 \times 10^{-5} \approx \frac{x^2}{0.00458}\) Now, we can solve for x: \(x = \sqrt{6.4 \times 10^{-5} \times 0.00458} \approx 0.000585 ~M\) Now that we have the value of x, we can find the concentrations and pH: [benzoic acid] = \((0.00458 - x) \approx 0.00458 ~M\) [benzoate ion] = \(x \approx 0.000585 ~M\) [hydrogen ion] = \(x \approx 0.000585 ~M\) Now, we can find the hydroxide ion concentration and pH: [hydroxide ion] = \(\frac{1.0 \times 10^{-14}}{[H^+]} = \frac{1.0 \times 10^{-14}}{0.000585} \approx 1.71 \times 10^{-11}~ M\) pH = \(-\log{[H^+]} = -\log{0.000585} \approx 3.23\)

Final results

We have now found all the required concentrations and the pH of the solution: [benzoic acid] = \(0.00458 ~M\) [benzoate ion] = \(0.000585 ~M\) [hydrogen ion] = \(0.000585 ~M\) [hydroxide ion] = \(1.71 \times 10^{-11}~ M\) pH = \(3.23\)

Key concepts

Acid Dissociation Constant

Understanding the acid dissociation constant, represented as K_a, is crucial for grasping the behavior of acids in water. K_a gives us insight into the strength of an acid; a higher value indicates a stronger acid. For benzoic acid, K_a is given as 6.4 × 10⁻⁵, suggesting that it is a weak acid because it partially dissociates in water.

In the example exercise, we use K_a to set up an equilibrium expression and calculate the extent of dissociation of benzoic acid. It involves using the initial concentration of the undissociated acid and the change in concentration upon reaching equilibrium. By knowing K_a and making some assumptions, such as that the dissociation is small relative to the initial concentration, we can simplify the equilibrium calculation and determine the concentrations of all species in solution.

pH Calculation

The pH of a solution is a measure of its acidity or alkalinity. Calculating pH is a fundamental skill in chemistry that involves converting hydrogen ion concentration to a logarithmic scale. In the context of the exercise, we calculate the pH by taking the negative logarithm of the concentration of hydrogen ions at equilibrium, pH = -log[H⁺].

For benzoic acid, after finding the concentration of hydrogen ions to be 0.000585 M, applying the pH formula gives us a pH of 3.23. This value indicates that the solution is acidic, which aligns with benzoic acid being a weak acid.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, leading to constant concentrations of the reactants and products. In the dissociation of benzoic acid, an equilibrium is established between the undissociated acid (C₆H₅CO₂H) and the ions (C₆H₅CO₂⁻ and H⁺).

In the exercise, we set up an equilibrium table to determine the concentrations of these species. The change in concentration of benzoic acid upon dissociation is represented by x, and by applying the equilibrium constant K_a, we solve for x to find the concentrations at equilibrium. It's important to note that equilibrium does not mean the reactants and products are in equal concentrations, but rather that their proportions do not change with time.

Molar Mass

The molar mass is the mass of one mole of a given substance, and it is a bridge between the macroscopic world of grams and the microscopic world of moles. In our exercise, the molar mass of benzoic acid (C₆H₅CO₂H) is 122.12 g/mol. We use this value to convert the mass of benzoic acid we have in grams (0.56 g) to moles before calculating its concentration in solution.

Simply put, the molar mass allows us to calculate how many molecules of benzoic acid were initially present in the water before any dissociation occurred. This initial concentration is the starting point for working out the equilibrium concentrations of all species in the solution.