What volume of 1.57 M HCl must be added to 0.240 L of 0.213 M dimethylamine to produce a buffer with a pH of 10.53?

Question

An acid–base buffer is a solution that contains a weak acid and its conjugate base. A buffer solution may be prepared by dissolving a weak base and then adding strong acid to obtain a buffer with the required pH.

What volume of 1.57 M HCl must be added to 0.240 L of 0.213 M dimethylamine to produce a buffer with a pH of 10.53? Dimethylamine has the formula (CH₃)₂NH, a K_b value of 5.89 ×× 10^–4, and a pK_b value of 3.23.

Answer 1

This will require the use of the Henderson-Hasselbalch equation.

pH = pKa + log [base]/[acid]

We are given the pKb, so the pKa of the conjugate acid is 14 – pKb = 14 – 3.23 = 10.77.

10.53 = 10.77 + log [base] / [acid].

The reaction of dimethylamine (DMA) with HCl is

DMA + HCl → DMAH¹⁺ + Cl^1-

So for every mole of HCl added is equal to the number of moles of DMAH¹⁺ produced.

The number of initial moles of DMA is 0.213 moles/L x 0.240 L = 0.0511 moles.

The moles DMA remaining after reaction with HCl can be expressed as 0.0511 – moles HCl added.

The moles HCl added can be expressed as moles HCl = moles DMAH¹⁺ produced.

The Henderson-Hasselbalch equation can now be written as

pH = 10.77 + log (0.0511 – moles HCl / moles HCl)

or 10.53 = 10.77 + log (0.0511 – moles HCl / moles HCl).

This can be solved for the ratio of base to acid.

Log ratio = 10.53 – 10.77 = -0.24 , ratio = 0.575

0.575 = (0.0511 – moles HCl) / moles HCl.

This can be solved for moles HCl moles HCl = 0.120 moles HCl.

Since the molarity of HCl to be added is 1.57 M, the #mL to be added can be found from the expression

Volume HCl = 0.120 moles x 1000 mL/L / 1.57 moles /L. Moles and L cancel leaving mL of HCl required = 76.4 mL.

Always check math.