Question

Complete the table below for the reaction of 20.0 mL of 0.150 M maleic acid with the indicated volume of 0.100 M OH^-. We will abbreviate maleic acid as H₂M. You must show your calculations to receive credit.

Please help, I don't know how to start. Thank you

(mL)	mmol of H2M	mmol of HM-	mmol of M2-	mmol of OH-
0.00
10.0
30.0
40.0
60.0
75.0

Answer 1

20.0 mL of 0.150 M maleic acid corresponds to

Initially, when no hydroxide ions are added, only 3.00 mmol of H2M are present.

When 10.0 mL of 0.100 M hydroxide ions are added, the number of mmol of hydroxide ions added

1.00 mmol of hydroxide ions will react with 1.00 mmol of H2M to form 1.00 mmol of HM-. mmol of H2M will remain unreacted. No amount of M2- or hydroxide ions is yet present.

When 30.0 mL of 0.100 M hydroxide ions are added, the number of mmol of hydroxide ions added

3.00 mmol of hydroxide ions will react with 3.00 mmol of H2M to form 3.00 mmol of HM- mmol of H2M will remain unreacted. No amount of M2- or hydroxide ions is yet present.

When 40.0 mL of 0.100 M hydroxide ions are added, the number of mmol of hydroxide ions added

4.00 mmol of hydroxide ions will react with 3.00 mmol of H2M to form 2.00 mmol of HM and 1.00 M2-. No amount of hydroxide ions is yet present.

When 60.0 mL of 0.100 M hydroxide ions are added, the number of mmol of hydroxide ions added

6.00 mmol of hydroxide ions will react with 3.00 mmol of H2M to form 3.00 mmol of M2-. No amount of hydroxide ions is yet present.

When 75.0 mL of 0.100 M hydroxide ions are added, the number of mmol of hydroxide ions added

7.50 mmol of hydroxide ions will react with 3.00 mmol of H2M to form 3.00 mmol of M2-. mmol of hydroxide ions is remaining.