Problem 65 - Chemistry: The Central Science

Question

Citric acid, which is present in citrus fruits, is a triprotic acid (Table 16.3). (a) Calculate the pH of a \(0.040 \mathrm{M}\) solution of citric acid. (b) Did you have to make any approximations or assumptions in completing your calculations? (c) Is the concentration of citrate ion \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{3-}\right)\) equal to, less than, or greater than the \(\mathrm{H}^{+}\)ion concentration?

Step-by-Step Solution

Answer

The pH of a 0.040 M citric acid solution is approximately 2.36. We made an assumption that the main contribution to the H+ concentration comes from the first dissociation step, and other dissociation steps can be neglected. The concentration of citrate ion (C6H5O73-) is negligible compared to the H+ ion concentration, and thus, less than the H+ ion concentration.

Step 1: Chemical reactions of citric acid dissociation

Citric acid (H3C6H5O7) is a triprotic acid, which means it can donate three protons (H+). The dissociation steps are as follows: 1st dissociation: \(H_3C_6H_5O_7 \rightleftharpoons H^+ + H_2C_6H_5O_7^-\) 2nd dissociation: \(H_2C_6H_5O_7^- \rightleftharpoons H^+ + HC_6H_5O_7^{2-}\) 3rd dissociation: \(HC_6H_5O_7^{2-} \rightleftharpoons H^+ + C_6H_5O_7^{3-}\)

Step 2: Find the Ka values of citric acid

The Ka values (acid dissociation constants) of the three dissociation steps are provided in Table 16.3: 1st dissociation: \(K_{a1} = 7.40 \times 10^{-4}\) 2nd dissociation: \(K_{a2} = 1.70 \times 10^{-5}\) 3rd dissociation: \(K_{a3} = 4.00 \times 10^{-7}\)

Step 3: Calculate the H+ concentration in the solution

Since the dissociation constants decrease through each step, most of the H+ contribution comes from the first dissociation. Therefore, we can make an assumption that the first dissociation mainly contributes to the H+ concentration and the contribution from the other dissociation steps can be neglected. We can apply the approximation with the first dissociation reaction and Ka1: \([H^+] [H_2C_6H_5O_7^{-}] = K_{a1}[H_3C_6H_5O_7]\) As the initial concentration of citric acid is 0.040 M, and assuming negligible change in concentration, we have: \([H^+](0.040 - [H^+]) \approx K_{a1}(0.040)\) Now, we can solve for [H+]: \([H^+] \approx \sqrt{K_{a1}(0.040)}\) \([H^+] \approx \sqrt{(7.40 \times 10^{-4})(0.040)}\) \([H^+] \approx 4.32 \times 10^{-3}\, M\)

Step 4: Calculate the pH of the solution

Now that we have the concentration of H+ ions, we can calculate the pH of the solution: \(pH = -\log[H^+]\) \(pH = -\log(4.32 \times 10^{-3})\) \(pH \approx 2.36\)

Step 5: Answer the additional questions

(a) The pH of a 0.040 M citric acid solution is approximately 2.36. (b) We made an assumption that the main contribution to the H+ concentration comes from the first dissociation step, and other dissociation steps can be neglected. (c) The concentration of citrate ion (C6H5O73-) is negligible compared to the H+ ion concentration. The third dissociation step has a Ka value of 4.00 × 10-7, meaning it's much less likely to occur. Thus, the concentration of citrate ion is less than the H+ ion concentration.

Key Concepts

Citric Acid Dissociation

Citric acid is an example of a triprotic acid, which means it can donate three protons (\( H^+ \)) in separate steps. This makes citric acid versatile and interesting in chemical reactions. The dissociation of citric acid occurs in three stages:

• First, \( H_3C_6H_5O_7 \rightarrow H^+ + H_2C_6H_5O_7^- \)
• Second, \( H_2C_6H_5O_7^- \rightarrow H^+ + HC_6H_5O_7^{2-} \)
• Finally, \( HC_6H_5O_7^{2-} \rightarrow H^+ + C_6H_5O_7^{3-} \)

The ability to donate more than one proton is key to understanding the behavior of citric acid in solutions. At each dissociation stage, fewer protons are released, and the corresponding anion forms.

Acid Dissociation Constants

For each dissociation step of citric acid, there is a specific acid dissociation constant, \( K_a \), that helps determine how readily the acid releases protons.

• The first dissociation constant, \( K_{a1} = 7.40 \times 10^{-4} \), is the most significant since it represents the primary proton donation.
• The second dissociation constant, \( K_{a2} = 1.70 \times 10^{-5} \), is much smaller, indicating reduced proton donation.
• The third dissociation constant, \( K_{a3} = 4.00 \times 10^{-7} \), is considerably smaller, showing minimal proton release in this stage.

These values decline progressively, reflecting the decreasing tendency for the acid to release protons as it continues to dissociate.

pH Calculation

To determine the pH of a citric acid solution, we mainly consider the first dissociation step due to its significant \( K_a \) value. This assumption simplifies calculations:

The formula we use is:

\([H^+] = \sqrt{K_{a1} \times [H_3C_6H_5O_7]}\)

Given \( [H_3C_6H_5O_7] = 0.040 \, M \), we find:

\([H^+] \approx \sqrt{(7.40 \times 10^{-4}) \times 0.040} \)
\([H^+] \approx 4.32 \times 10^{-3} \, M \)

The pH is then calculated using:

\( pH = -\log[H^+] \approx -\log(4.32 \times 10^{-3}) \approx 2.36 \)

This pH indicates a relatively acidic solution, typical of citric acid.

Ion Concentration Comparison

In comparing ion concentrations, we focus on proton concentration \( [H^+] \) and the concentration of the citrate ion ( \( C_6H_5O_7^{3-} \)).

The significant drop in \( K_a \) values between dissociation steps implies:

• The first dissociation contributes most to \( [H^+] \)
• The concentration of \( C_6H_5O_7^{3-} \) is minimal, due to the low \( K_{a3} \) indicating it rarely forms

Hence, \( [C_6H_5O_7^{3-}] \) is considerably less than \( [H^+] \), confirming that most citric acid molecules do not reach the third dissociation stage. This analysis helps in understanding the minor presence of fully dissociated ions in a solution.