Question

A 25.00 mL aliquot of a 0.100 M maleic acid, HOOC-CH=CH-COOH (pK1=1.87) and (pK2 = 6.23) was titrated with 0.1000 M NaOH. Construct a titration curve by calculating pH at a) initial pH b) after adding 5.00 mL of NaOH c) after adding 24.90 mL of NaOH d) at equivalence point e) after adding 25.01 mL of NaOH f) after adding 25.50 ml of NaOH g) after adding 49.90 mL NaOH h) after adding 50.00 mL NaOH i) after adding 50.01 mL NaOH j) after adding 51.00 mL of NaOH

Answer 1

This is a very long question, however, I will give you some guidance to do this.

First we need to calculate the equivalence volume at the equivalence point:
Vb = 0.1 * 25 / 0.1 = 25 mL

then as is a diprotic acid it would be 2*Vb = 50 mL.

Innitial pH.

The reaction is along with the moles and ICE chart):
moles acid = 0.1 * 0.025 = 2.5x10^-3 moles

r: H₂C₄H₂O₄ ----------> HC₄H₂O₄^- + H⁺   Ka1 = 10^-1.87 = 1.35x10^-2
i. 0.1 0 0
e. 0.1-x x x

1.35x10^-2 = x² / 0.1-x
1.35x10^-2(0.1-x) = x²
1.35x10^-3 - 1.35x10^-2x = x²
x² + 1.35x10^-2x- 1.35x10^-3 = 0

Using the quadratic formula, we get the following x value:
x = 0.0306 M = [H⁺]
pH = -log(0.0306) = 1.51

When you add 5 mL of base, the moles added are:
moles OH = 0.1 * 0.005 = 0.0005 moles

r: H₂C₄H₂O₄ + OH^-----------> HC₄H₂O₄^- + H₂O

moles of acid remaining = 0.0025 - 0.0005 = 0.0020 moles

Now we use the HH equation:
pH = pKa + log(A-/HA)
pH = 1.87 + log(0.002/0.0025)
pH = 1.77

Now try to follow these procedure to get the other pH

at equivalence point is
pH = pK1 + pK2 / 2
pH = 1.87 + 6.23 / 2 = 2.05

Hope this guidance helps