Question

A biochemist has 100 mL of a buffered solution at pH 7.3. The concentration of the buffer is 0.1 M, and the pKa of the buffer is 7.8. If the biochemist adds 80 mL of a 0.1 N KOH solution to the above buffered solution, what will be the final pH to the nearest tenth of a unit?

Answer 1

We have Henderson equation

pH = pka + log [conjugate base] / [acid]

Let acid be HA , its conjugate base is A- , pka of HA = 7.8 , pH = 7.3

we substitute all these in Henderson eq

7.3 = 7.8 +log [A-]/[HA]

[A-] = 0.31623 [HA] ...............(1)

given concentration of buffer = 0.1 M

hence [HA] + [A-] = 0.1 .................(2)

by solving (1) ( 2) we get

[HA] = 0.076 ,   [A-] = 0.024 M

volume =100ml = 0.1 L

Moles of HA = M x V = 0.076 x 0.1 = 0.0076

Moles of A- = M x V = 0.024 x 0.1 = 0.0024

Moles of KOH added = M x V = 0.1 x ( 80/1000) = 0.008

KOH gives OH- which reacts with HA to give A- ,  

HA + OH- ---> H2O (l) + A-

OH- moles reacted = HA moles = 0.0076

Thus excess KOH left after reacting with 0.0076 moles HA = 0.008-0.0076 = 0.0004

total solution volume = 180 ml = 0.18L

[OH-] = moles/volume = 0.0004/0.18 = 0.0022 L

pOH = -log [OH-] = -log ( 0.0022) = 2.65

pH = 14-pOH = 14-2.65 = 11.35

( please recheck values given i.e volume of KOH etc , if any corrections present notify it )