Buffer chemistry

Question

How many mL of 1.07 M HCIO4 (aq) should be added to 1.00 g of imidazole to give it a pH of 6.993?

Answer 1

Imidazole (C₃N₂H₄) is a weak base. We can represent this simply as B

B + HClO₄ ==> BH⁺ + ClO₄^-

If we use a modification of the Henderson Hasselbalch equation, we can solve for BH⁺ and then for HClO₄

pOH = pKb + log [BH⁺] / [B]

pOH = 14 - pH = 14 - 6.993 = 7.007

pKb imidazole = 7.05 (looked it up)

7.007 = 7.05 + log [BH⁺] / [B]

log [BH⁺] / [B] = -0.043

[BH⁺] / [B] = 0.906

molar mass imidazole (C₃N₂H₄) = 68.1 g/mol

1.00 g x 1 mol / 68.1 g = 0.0147 mols C₃N₂H₄

[BH⁺] / [B] = 0.906

B + HClO₄ ==> BH⁺

0.0147.....0............0

-x...........+x...........+x

0.0147-x...x...........x

[BH⁺] / [B] = 0.906

x / 0.0147-x = 0.906

x = BH⁺ = 0.00699 moles = 0.00699 moles HClO₄

mls of 1.07 M HClO₄ = 0.00699 mols HClO4 x 1 L / 1.07 moles x 1000 ml / L = 6.53 mls HClO₄

(please be sure to check all of the math)