Question

Consider a \(0.45 \mathrm{M}\) solution of ascorbic acid, \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6},\) one form of Vitamin C. $$ \begin{array}{ll} \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}(a q) \leftrightharpoons \mathrm{HC}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^{-}(a q)+\mathrm{H}^{+}(a q) & K_{\mathrm{a}}=7.9 \times 10^{-5} \\ \mathrm{HC}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^{-}(a q) \leftrightharpoons \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^{2-}(a q)+\mathrm{H}^{+}(a q) & K_{\mathrm{a}}=1.6 \times 10^{-12} \end{array} $$ Calculate the \(\mathrm{pH}\) of this solution and estimate \(\left[\mathrm{HC}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^{-}\right]\) and \(\left[\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^{2-}\right]\)

Short answer

Based on the given dissociation reactions and Ka values, the following results are obtained: 1. The pH of the ascorbic acid solution is approximately 2.396. 2. The concentration of the first dissociated form, HC6H6O6-, is approximately 0.00403 M. 3. The concentration of the second dissociated form, C6H6O62-, is approximately 1.6 x 10^-12 M.

Step-by-step solution

Find the initial concentration of H+ ions

Initially, the ascorbic acid molecule is not dissociated. Therefore, the concentration of \(\mathrm{H}^+\) ions is \(0\mathrm{M}\).

Set up the first equilibrium expression

From the first dissociation reaction, $$ \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6} \leftrightharpoons \mathrm{HC}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^{-} + \mathrm{H}^{+}$$ The K\(_a\) expression is: $$K_a = \frac{[\mathrm{HC}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^-][\mathrm{H}^+]}{[\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}]}$$ Now, let \(x\) be the concentration of the dissociated \(\mathrm{HC}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^-\) and \(\mathrm{H}^+\) ions. The concentration of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}\) will be \(0.45 - x\). $$7.9 \times 10^{-5} = \frac{x \times x}{0.45-x}$$

Solve for x (first dissociation)

As the given \(K_a\) value for the first dissociation reaction is quite small, x will also be quite small, allowing us to assume that \(0.45 - x \approx 0.45\). Hence: $$7.9 \times 10^{-5} = \frac{x^2}{0.45}$$ $$x = 0.00403$$ This means that the concentrations of \(\mathrm{HC}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^-\) and \(\mathrm{H}^+\) ions are approximately equal to \(0.00403\mathrm{M}\) after the first dissociation.

Set up the second equilibrium expression

From the second dissociation reaction, $$ \mathrm{HC}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^{-} \leftrightharpoons \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^{2-} + \mathrm{H}^{+}$$ Now, let \(y\) be the concentration of the dissociated \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^{2-}\) and additional concentration of \(\mathrm{H}^+\) ions. The concentration of \(\mathrm{HC}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^-\) will be \(0.00403 - y\). The second K\(_a\) expression is: $$K_a = \frac{[\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^{2-}][\mathrm{H}^+]}{[\mathrm{HC}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^-]}$$ Replacing the given K\(_a\) and variables with their values: $$1.6 \times 10^{-12} = \frac{y \times (0.00403+y)}{0.00403-y}$$

Solve for y (second dissociation)

The dissociation constant of the second reaction is much smaller than the first. Thus, we can assume that \(y\) is very small compared to \(0.00403\). This assumption allows us to write: $$1.6 \times 10^{-12} = \frac{y \times 0.00403}{0.00403}$$ $$y = 1.6 \times 10^{-12}$$ This is the concentration of \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^{2-}\).

Calculate the pH of the solution

The final concentration of \(\mathrm{H}^+\) ions is approximately the sum the individual concentrations from the first and second dissociation reactions. $$[\mathrm{H}^+] = 0.00403 + 1.6 \times 10^{-12} \approx 0.00403$$ Now, we can calculate the pH as: $$pH = -\log_{10}[\mathrm{H}^+] = -\log_{10}(0.00403) = 2.396$$ #Results# 1. The pH of the solution is approximately 2.396. 2. The concentration of \(\mathrm{HC}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^-\) is approximately 0.00403 M. 3. The concentration of \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}^{2-}\) is approximately \(1.6 \times 10^{-12}\mathrm{M}\).

Key concepts

Acid Dissociation Constant

The acid dissociation constant, represented as Ka, is a quantitative measure of an acid's strength. In simpler terms, it tells us how well an acid can donate a proton (H+) in solution. The higher the Ka value, the stronger the acid, meaning it dissociates more completely. For example, the ascorbic acid in our exercise dissociates in water to form hydrogen ions (H+) and the ascorbate ion.

Understanding Ka is crucial for predicting the behavior of acids in solutions. It can be calculated using the equilibrium concentrations of the reactants and products. The expression for ascorbic acid's first dissociation step is given by: \[ K_a = \frac{[\mathrm{HC}_6 \mathrm{H}_6 \mathrm{O}_6^-][\mathrm{H}^+]}{[\mathrm{H}_2 \mathrm{C}_6 \mathrm{H}_6 \mathrm{O}_6]} \]
Here, the square brackets denote concentration, and the products are on the numerator while the reactant is on the denominator. A concept that often accompanies Ka is the pKa, which is simply the negative logarithm of Ka and provides a more convenient way to express acid strength, especially for very weak or very strong acids.

pH Calculation

The pH of a solution is a measure of its acidity or basicity. In technical terms, it is the negative logarithm of the hydrogen ion concentration. This seemingly simple scale tells us a lot about the solution's chemical nature. A pH less than 7 indicates acidity, while a pH greater than 7 indicates basicity. At pH 7, the solution is neutral.

To calculate pH, the hydrogen ion concentration needs to be found first. This is where the acid dissociation constant comes into play. For ascorbic acid, we use the Ka to find the concentration of hydrogen ions produced in the dissociation reaction. Once we have the hydrogen ion concentration (denoted as [H+]), the pH is found using the formula: \[ pH = -\log_{10}[\mathrm{H}^+] \]
Therefore, the pH is found to be a logarithmic function of the hydrogen ion concentration, allowing us to quickly and easily determine the acidity of almost any aqueous solution.

Chemical Equilibrium

Chemical equilibrium refers to the state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. As a result, the concentrations of reactants and products remain constant over time. This doesn't mean the reactants and products are in equal concentrations, but rather that their rates of formation are stable.

For ascorbic acid, equilibrium is reached when it partially dissociates in water. We describe this equilibrium using equilibrium constants, such as Ka, which reflects the ratio of product concentrations to reactant concentrations at equilibrium. By setting up the equilibrium expressions for the dissociation of ascorbic acid, we can solve for the unknown concentrations, allowing us to calculate pH and understand solution behavior at equilibrium.

Moreover, the concept of chemical equilibrium is vital when considering the reactions' reversibility—a fundamental aspect in predicting the outcome of chemical processes and understanding reaction dynamics in solutions.