Question

When 10.0 mL of 0.10 M HCOOH is titrated with .10 M NaOH, at the equivalence point of the pH the solution will be:

A) Greater than 7 B) 7.00 C) Less than 7

Answer 1

Calculate moles of acid present:

(0.10 mol/L) (0.01 L) = 1 x 10^-3 moles

Determine moles of base required to react equivalence point:

CH₃COOH + NaOH ---> CH₃COONa + H₂O

There is a 1:1 molar ratio between formic acid and sodium hydroxide.

Therefore, 1 x 10^-3 moles of base required

Calculate volume of base solution required:

1 x 10^-3 mol divided by 0.10 mol/L = 0.01 L = 10 mL

Total volume of the solution = 10 + 10 mL = 0.02L

The solution is now completely composed of a salt of a weak acid. The pH of this solution will be basic.

1) Calculate molarity of sodium formate:

1 x 10^-3 mol / 0.02 L = 0.05 M

2) Calculate the K_b of sodium acetate:

K_w = K_aK_b

1.00 x 10^-14 = (1.8 x 10^-4) (x)

x = 5.55 x 10^-11

3) Calculate pH of the solution:

5.55 x 10^-11 = [(x) (x)] / 0.05

x = 1.67 x 10^-6 M (this is the hydroxide ion concentration)

pOH = 5.778

pH = 14 - pOH = 8.222

So Answer is A

pH is greater than 7