Question

You are given a buffer that contains a weak base whose concentration is 0.600 M and its conjugate weak acid whose concentration is 0.260 M. The volume of this solution is 0.375 L. If 0.080 L of an HCl solution with a concentration of 1.10 M is added to this solution, what will the pH of the combined solution be? The pKa of the weak acid is 4.67.

Answer 1

4.54

Explanation

No of mole of weak base = (0.600mol/1L)×0.375L= 0.225mol

No of mole of conjucate acid = (0.260mol/1L)×375L = 0.0975mol

No of mole of HCl added = (1.10mol/1L)×0.080L = 0.088mol

HCl react with weak base

HCl + B - - - -> BH+ + Cl-

Stoichiometrically, 1mole of HCl react with 1mole of Base to give 1mole of Conjucate acid

0.088mole of HCl react with 0.088mole of Base to produce 0.088mole of conjucate acid

After addition of HCl

No of mole of Base = 0.225mol - 0.088 = 0.137moles

No of mole of Acid = 0.0975 + 0.088 = 0.1855moles

Total volume = 0.375L + 0.080L = 0.455L

[Base] = (0.137moles/0.455L) ×1L =0.3011M

[Conjucate acid] = (0.1855mol/0.455L)×1L =0.4077M

Henderson-Hasselbalch equation is

pH= pKa + log([Base] / [Conjucate acid])

= 4.67 + log(0.3011M/0.4077M)

= 4.67 - 13

= 4.54

=