the titration of 25.0 mL of a 0.040 M aqueous solution of maleic acid C2H2(COOH)2, with aqueous 0.10 M NaOH solution. For maleic acid, Ka1 = 0.013 and Ka2 = 8.5 x 10-7

How many milliliters of the 0.10 M NaOH solution are needed to completely react all the maleic acid in 25.0 mL of a 0.040 M solution of maleic acid?

If we write maleic acid as H2M, then

H2M + 2NaOH ==> Na2M + 2H2O
mols H2M = M x L
\mols NaOH = twice that (look a the coefficients)
\Then M NaOH = mols NaOH/LNaOH
You know M and mols, solve for L NaOH and convert to mL.

To solve this problem, we will use the concept of acid-base titration. In an acid-base titration, a base of known concentration is added to an acid solution until the reaction is complete.

To find the number of milliliters of the NaOH solution needed to completely react all the maleic acid present in 25.0 mL of a 0.040 M solution, we can follow these steps:

Step 1: Determine the balanced chemical equation for the reaction between maleic acid (C2H2(COOH)2) and NaOH (sodium hydroxide).

The balanced chemical equation for the reaction between maleic acid and NaOH is:

C2H2(COOH)2 + 2 NaOH → C2H2(COONa)2 + 2 H2O

Step 2: Convert the volume of the maleic acid solution from milliliters to liters.

We have 25.0 mL of maleic acid solution, which is equivalent to 0.025 L (since 1 L = 1000 mL).

Step 3: Calculate the amount of maleic acid (in moles) present in the 25.0 mL solution.

To do this, we use the formula:

Amount (in moles) = Concentration (in M) × Volume (in L)

Amount of maleic acid = 0.040 M × 0.025 L = 0.001 mol

Step 4: Use stoichiometry to find the amount of NaOH required to react with the maleic acid.

From the balanced chemical equation, we can see that 1 mole of maleic acid reacts with 2 moles of NaOH. Therefore, the moles of NaOH required will be twice the moles of maleic acid.

Amount of NaOH = 2 × 0.001 mol = 0.002 mol

Step 5: Convert the amount of NaOH from moles to milliliters.

To do this, use the formula:

Volume (in mL) = Amount (in moles) ÷ Concentration (in M)

Volume of NaOH = 0.002 mol ÷ 0.10 M = 0.020 L (since 1 L = 1000 mL)

Volume of NaOH = 0.020 L × 1000 mL/L = 20 mL

Therefore, it will take 20 milliliters of the 0.10 M NaOH solution to completely react all the maleic acid in 25.0 mL of a 0.040 M solution of maleic acid.