Question

A compound has carboxyl group, an aminium group, and a guanidinium group. The pKas of the groups are 3.6, 9.7, and 12.8, respectively. A biochemist has 100 mL of a 0.10 M solution of this compound at a pH of 9.2. She adds 37 mL of 0.1 M HCl. To the nearest hundredth of a unit, what will be the pH of the solution after addition of the HCl (assume full stirring to reach a new equilibrium)?

Answer 1

Sol:-

Let the given tribasic acid is H₃A. Now the equation for H₂A^- acid at equilibrium will be :

H₂A^- <----> H⁺ + HA^2-

According to Henderson-Hasselbalch equation , we have

9.2 = 9.7 + log [H₂A^-]/[HA^2-]

log [H₂A^-]/[HA^2-] = 9.2 - 9.7

log [H₂A^-]/[HA^2-] = - 0.5

[H₂A^-]/[HA^2-] = 10^{- 0.5}

[H₂A^-]/[HA^2-] = 0.316

After the addition of HCl

moles of compound = 0.100 M x 0.100 L = 0.0100 mol

so

moles of HA^2- + moles of H₂A^- = 0.0100 mol .........(1)

and mol HA^2-/mol H₂A^- = 0.316

mol HA^2- = 0.316 mol H₂A^- ..................(2)

from equation (1) and (2) , we have

0.316 molesH₂A^- + moles of H₂A^- = 0.0100 mol

1.316 moles of H₂A^- = 0.0100 mol

moles of H₂A^- = 0.0100 / 1.316

moles of H₂A^- = 0.0076 and

moles HA^2- = 0.0024 mol

moles HCl added = 0.10 M x 0.037 L = 0.0037 mol HCl

Moles of H₂A^- = 0.0024

when HCl reacts with H₂A^-

we have unreacted 0.0013 moles of HCl

0.0013 moles of HCl converts 0.0013 moles of H₂A^- into H₃A

so

Moles of H₃A = 0.0013 mol

Moles of H₂A ^- = 0.001 + 0.0076 = 0.0086

H3A <----> H⁺ + H₂A^- , pKa = 1.8

where mol H₃A is 0.0013 mol and mol H₂A^- is 0.0086

again by using Henderson-Hasselebalch equation , we have

pH = 3.7 + log(0.0086/0.0013)

pH = 3.7 + log 6.615

pH = 3.7 + 0.82

pH = 4.52