Answer:
pH = 3.19
Step-by-step explanation:
Citric acid is a triprotic acid, meaning that it can donate three protons (H+ ions) in solution. Each proton has its own acid dissociation constant (Ka). For citric acid, the three Ka values are:
Ka1 = 7.4 × 10^-4
Ka2 = 1.7 × 10^-5
Ka3 = 4.0 × 10^-7
Assuming that the citric acid is fully ionized, the initial concentration of each ion can be calculated from the initial concentration of citric acid:
[H3Cit] = [CitH2-] = [CitH3-] = 2.55 × 10^-2 M
The total concentration of H+ ions in solution is the sum of the concentrations of each ion:
[H+] = [H3Cit] + 2[H2Cit-] + 3[Cit^2-]
We can use the Ka values to calculate the concentrations of the ions:
[H+][CitH3-] / [H3Cit] = Ka1
[CitH2-][H+] / [CitH3-] = Ka2
[Cit^2-][H+] / [CitH2-] = Ka3
Rearranging each equation to solve for [H+], and substituting in the concentrations of the ions, we get:
[H+] = √(Ka1Ka2Ka3 / [H3Cit](1 + Ka1/[H+] + Ka1Ka2/([H+]^2)))
Plugging in the values, we get:
[H+] = √(7.4 × 10^-4 × 1.7 × 10^-5 × 4.0 × 10^-7 / (2.55 × 10^-2) × (1 + 7.4 × 10^-4/[H+] + 7.4 × 10^-4 × 1.7 × 10^-5/([H+]^2)))
Solving this equation numerically, we get:
[H+] = 6.50 × 10^-4 M
Therefore, the pH of the solution is:
pH = -log[H+]
pH = -log(6.50 × 10^-4)
pH = 3.19
So the original pH of the triprotic citric acid solution is approximately 3.19.