Question

Using the equations (discussed above) showing the relationship between [acid], [conjugate base] and [buffer concentration], mathematically show the negligible change in pH even when 2µL of 10M HCl is added to 100 ml of 50 mM Phosphate buffer at pH 6.9. To address this problem, the following steps will be needed: (i) You will have to calculate the pH of the buffer after adding the 2µL of 10M HCl (ii) To do this, calculate the new concentration of [acid] and [conjugate base] after the addition of 2µL of 10M HCl, and (iii) to appreciate the effect of buffers, show the pH change in 100 ml of water after upon adding 2µL of 10M HCl that does not contain the phosphate buffer ions.

Answer 1

PH = PKa + log [A^- ]/[HA] (Henderson-Hasselbalch equation)

Henderson-Hasselbalch equation can be used for pH calculation of a solution containing pair of acid and conjugate base.

Convert 2µL of HCl to ml

2µL = 0.002ml

New concentration of HCl can be calculated by using formula M₁V₁ = M₂V₂

0.002 x 10 = 100 M₂

M₂ = 0.002 x 10/100 = 0.0002M

ii) To calculate the PH when 2µL of 10M HCl is added to 100 ml of 50 mM Phosphate buffer at pH 6.9, first calculate the new concentration of [acid] and [conjugate base] after the addition of 2µL of 10M HCl,

Phosphate buffer at pH 6.9, so.(pKa2 = 7.2) is taken

100 ml of 50 mM Phosphate buffer = 0.025

The new concentration would be:         

A^- = 0.025 – 0.0001 = 0.0249

[HA] = 0.025 + 0.0001 = 0.0251

i) PH = 7.2 + log (0.0249/0.0251)

PH = 7.2 – 00347 = 7.196

iii) PH change in 100 ml of water after upon adding 2µL of 10M HCl that does not contain the phosphate buffer ions.

HCl = 0.0002M

The HCl is a strong acid and is 100% ionized in water.

The hydronium ion concentration is 0.0002 M.

PH = - log (0.0002) = 3.7