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The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (HOOCCOOH) and must be removed before the stems are used to make rhubarb pie. If pKa1 = 1.23 and pKa2 = 4.19, what is the pH of a 0.0335 M solution of oxalic acid?

User Iasonas
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Answer:

The pH of a 0.0335 M solution of oxalic acid is approximately 1.214.

Step-by-step explanation:

To find the pH of a 0.0335 M solution of oxalic acid, we need to consider the dissociation of the acid into its respective ions. Oxalic acid (HOOCCOOH) is a diprotic acid, meaning it can donate two protons (H+ ions) in solution. The dissociation reactions are as follows:

1. HOOCCOOH ⇌ H+ + OOCCOOH

2. OOCCOOH ⇌ H+ + OOCOO-

The acid dissociation constant, Ka, is related to the concentration of the ions in solution. The pKa value is the negative logarithm of the Ka value.

Given pKa1 = 1.23 and pKa2 = 4.19, we can use these values to calculate the pH of the solution:

1. Calculate the concentration of H+ ions from the first dissociation:

[H+] = 10^(-pKa1) * [HOOCCOOH]

= 10^(-1.23) * 0.0335

= 0.0603 M

2. Calculate the concentration of H+ ions from the second dissociation:

[H+] = 10^(-pKa2) * [OOCCOOH]

= 10^(-4.19) * 0.0335

= 0.000682 M

3. Calculate the total concentration of H+ ions in solution:

[H+]total = [H+] from the first dissociation + [H+] from the second dissociation

= 0.0603 + 0.000682

= 0.06098 M

4. Calculate the pH using the concentration of H+ ions:

pH = -log10[H+]

= -log10(0.06098)

≈ 1.214

Therefore, the pH of a 0.0335 M solution of oxalic acid is approximately 1.214.

User David Mertens
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