Question

Consider a \(0.33 \mathrm{M}\) solution of the diprotic acid \(\mathrm{H}_{2} \mathrm{X}\). $$ \begin{array}{ll} \mathrm{H}_{2} \mathrm{X} \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HX}^{-}(a q) & K_{\mathrm{a} 1}=3.3 \times 10^{-4} \\ \mathrm{HX}^{-} \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{X}^{2-}(a q) & K_{\mathrm{a} 2}=9.7 \times 10^{-8} \end{array} $$ Calculate the \(\mathrm{pH}\) of the solution and estimate \(\left[\mathrm{HX}^{-}\right]\) and \(\left[\mathrm{X}^{2-}\right]\)

Short answer

To summarize, the pH of the given diprotic acid \(\mathrm{H}_{2}\mathrm{X}\) solution is approximately 2. The concentration of \(\left[\mathrm{HX}^{-}\right]\) is approximately \(0.00999 \mathrm{M}\), and the concentration of \(\left[\mathrm{X}^{2-}\right]\) is approximately \(9.7 \times 10^{-8} \mathrm{M}\).

Step-by-step solution

Calculate the concentration of \(\mathrm{H}^{+}\) using \(K_{\mathrm{a} 1}\)

For the first dissociation reaction, we have the equilibrium constant \(K_{\mathrm{a} 1} = 3.3 \times 10^{-4}\). We can set up the equilibrium expression for the reaction: $$K_{\mathrm{a} 1} = \frac{[\mathrm{H}^+][\mathrm{HX}^-]}{[\mathrm{H}_{2}\mathrm{X}]}$$ We know that \([\mathrm{H}_{2}\mathrm{X}] = 0.33 \mathrm{M}\) (initial concentration). Let's use \(x\) for the change in concentration of \(\mathrm{H}^+\) during the reaction, so then the change in concentration of \(\mathrm{HX}^-\) will also be \(x\), and the change in concentration of \(\mathrm{H}_{2}\mathrm{X}\) will be \(-x\). Therefore, at equilibrium we have: $$[\mathrm{H}^+] = x$$ $$[\mathrm{HX}^-] = x$$ $$[\mathrm{H}_{2}\mathrm{X}] = 0.33 - x$$ We can substitute these values into the equilibrium expression and solve for \(\mathrm{H}^+\): $$3.3 \times 10^{-4} = \frac{x \cdot x}{(0.33 - x)}$$

Solve for the concentration of \(\mathrm{H}^{+}\)

Considering that \(K_{\mathrm{a} 1}\) is a relatively small value, it suggests that only a small portion of the \(\mathrm{H}_{2}\mathrm{X}\) dissociates. We can assume that \(x\) is very small in comparison to \(0.33\), which simplifies the equation: $$3.3 \times 10^{-4} \approx \frac{x^2}{0.33}$$ $$x^2 \approx 0.33 \cdot 3.3 \times 10^{-4}$$ $$x = \sqrt{0.33 \cdot 3.3 \times 10^{-4}}$$ $$x \approx 0.00999$$ Therefore, the concentration of \(\mathrm{H}^+\) is approximately \(0.00999 \mathrm{M}\).

Calculate the pH of the solution

Now that we have the concentration of \(\mathrm{H}^+\), we can calculate the pH of the solution by using the pH formula: $$\mathrm{pH} = -\log_{10} [\mathrm{H}^+]$$ $$\mathrm{pH} \approx -\log_{10} (0.00999)$$ $$\mathrm{pH} \approx 2$$ The pH of the solution is approximately 2.

Estimate the concentration of \(\left[\mathrm{HX}^{-}\right]\) and \(\left[\mathrm{X}^{2-}\right]\)

Since we already know that the concentration of \(\mathrm{HX}^-\) is equal to the concentration of \(\mathrm{H}^+\), we can directly estimate that \([\mathrm{HX}^-] \approx 0.00999 \mathrm{M}\). Now, for the second dissociation reaction, we have the equilibrium constant \(K_{\mathrm{a} 2} = 9.7 \times 10^{-8}\). We can set up the equilibrium expression for the reaction: $$K_{\mathrm{a} 2} = \frac{[\mathrm{H}^+][\mathrm{X}^{2-}]}{[\mathrm{HX}^-]}$$ We know that: \([\mathrm{HX}^-] \approx 0.00999 \mathrm{M}\) \([\mathrm{H}^{+}] \approx 0.00999 \mathrm{M}\) Now, let's use \(y\) for the change in concentration of \(\mathrm{X}^{2-}\) during the reaction. Then, at equilibrium we have: $$[\mathrm{X}^{2-}] = y$$ Substitute these values into the equilibrium expression and solve for \(y\) (concentration of \(\mathrm{X}^{2-}\)): $$9.7 \times 10^{-8} = \frac{0.00999 \cdot y}{0.00999}$$ $$y = 9.7 \times 10^{-8}$$ Therefore, the concentration of \(\mathrm{X}^{2-}\) is approximately \(9.7 \times 10^{-8} \mathrm{M}\).

Key concepts

pH Calculation

Understanding how to calculate the pH of a solution is crucial when studying acid-base chemistry. The pH is a measure of the acidity or basicity of an aqueous solution. It is defined as the negative logarithm (base 10) of the hydrogen ion concentration, expressed in moles per liter (Molarity).
For any solution where the hydrogen ion concentration, \( [H^+] \), is known, the pH can be calculated using the following formula:
\[ pH = -\log_{10} [H^+] \]
In the context of a diprotic acid, which can donate two protons (hydrogen ions), it's essential to consider both dissociation steps when calculating the pH. For the first dissociation, the \( [H^+] \) generated from the disassociation must be accurately determined, allowing us to then use the above formula to find the pH of the solution. An approximation that simplifies calculations is that \( [H^+] \) is only considered for the first dissociation, because in most cases, the second dissociation contributes significantly less to the overall \( [H^+] \). This is due to the second dissociation constant, \( K_{a2} \), usually having a much smaller value compared to the first dissociation constant, \( K_{a1} \).

Equilibrium Constant

The equilibrium constant, denoted by \( K \), is a value that expresses the ratio of the concentrations of the products of a reaction to the concentrations of the reactants, each raised to the power of their stoichiometric coefficients, at equilibrium. For an acid dissociation reaction, which is a type of equilibrium, we specifically refer to this constant as the acid dissociation constant (Ka).
These constants are crucial in predicting the extent of a reaction, understanding reaction dynamics, and calculating concentrations of reactants and products at equilibrium. The greater the value of \( K \) (or \( Ka \) for acids), the more the reaction favors the formation of products, meaning that the acid is stronger and dissociates more in water. Conversely, a smaller \( K \) value indicates less dissociation and a weaker acid.
When we encounter a diprotic acid, it's important to consider that there are two equilibrium constants, \( K_{a1} \) for the first dissociation, and \( K_{a2} \) for the second. Typically, \( K_{a1} > K_{a2} \), indicating that the first proton is dissociated more readily than the second.

Acid Dissociation Constant

The acid dissociation constant, \( Ka \), is specific to the context of acid-base reactions, providing a quantitative measure of the strength of an acid in solution. Each acid has its \( Ka \), and for diprotic acids such as \( H_2X \), there are two dissociation constants to consider: \( K_{a1} \) and \( K_{a2} \).
\[ K_{a1} = \frac{[H^+][HX^-]}{[H_{2}X]} \]
\[ K_{a2} = \frac{[H^+][X^{2-}]}{[HX^-]} \]
In equations representing dissociation, \( Ka \) portrays the ratio of the concentration of the dissociated (ionized) form of the acid to the concentration of the undissociated (non-ionized) form. Because dissociation is reversible, these constants are derived at equilibrium state of the reaction.
The magnitude of \( Ka \) reflects the acid's tendency to lose a proton; a high \( Ka \) signifies a strong acid that dissociates completely in solution, whereas a low \( Ka \) indicates a weak acid that remains mostly undissociated. Knowing the \( Ka \) values allows us to predict the position of equilibrium and calculate the concentrations of the various species in solution, which is essential in the pH calculation of acid solutions.