Question

A 20.00 mL sample of a diprotic acid is titrated with a solution of sodium hydroxide (NaOH). The balanced chemical equation for the reaction is:

\[ \text{H}_2A (aq) + 2\text{NaOH} (aq) \rightarrow \text{Na}_2A (aq) + 2\text{H}_2O (l) \]
If it takes 25.00 mL of 0.10 M NaOH solution to reach the equivalence point, what is the concentration of the diprotic acid (H₂A) in mol/L?
A) 0.05 M
B) 0.10 M
C) 0.20 M
D) 0.40 M

Answer 1

The concentration of the diprotic acid (H2A) is calculated to be 0.0625 M by dividing the moles of acid by the volume of the acid solution. However, this result does not match any of the given options, indicating a possible typo in the question.

Step-by-step explanation:

To calculate the concentration of the diprotic acid (H2A) in mol/L using titration data, we can apply stoichiometry. The balanced chemical equation given is:

H2A (aq) + 2NaOH (aq) → Na2A (aq) + 2H2O (l)

We know that it takes 25.00 mL of 0.10 M NaOH to neutralize the acid, which means there are:

25.00 mL × 0.10 mol/L = 0.0025 mol of NaOH

Since the ratio of NaOH to H2A is 2:1, there must be half as many moles of H2A, or 0.00125 mol of H2A. The volume of the acid sample is 20.00 mL (or 0.02000 L) so the molarity of H2A is:

0.00125 mol ÷ 0.02000 L = 0.0625 mol/L

However, this is not one of the options provided (A, B, C, or D). Since this may be due to a typo in the given options, I would advise you to double-check your question and the available answers. If 0.0625 M was meant to be among the choices, then that would be the correct concentration of the diprotic acid H2A.