Question

Please use a diagram to explain!

CE U. TU I TIUI. The 13. Joe Student is titrating two weak acids with KOH solution. They have the same original concentration, but acid A is 10% dissociated while acid B is stronger and is 20% dissociated. Joe therefore expects acid B to require twice the volume of KOH solution that acid A requires. However, it doesn't work out that way. What is wrong with Joe's reasoning?

Answer 1

13.

Let us represent two weak acids as HA and HB. They react with KOH

HA+KOH → KA+H2O HB + KOH → KB + H2O

When HA reacts with KOH, the number of moles of KOH used depends on the number of moles of HA and not on number of moles of hydronium ions present. In other words, the initial concentration of HA (and HB) matters and percent dissociation does not matters.

In other words, total concentration of HA (or HB) matters which includes dissociated and undissociated form.

This is explained with the help of following diagram.

1 L of 1 M HA 1 L of 1 M HB 20% 10% dissociation dissociation Out of 1 mole of HA present, 90% or 0.9 mole is undissociated HA and the remaining 10% or 0.1 mole is dissociated Out of 1 mole of HB present, 80% or 0.8 mole is undissociated HB and the remaining 20% or 0.2 mole is dissociated But number of moles of KOH required depends on total number of moles of acid, that includes 0.9 moles undissociated and 0.1 mole dissociated. 1 고고 But number of moles of KOH required depends on total number of moles of acid, that includes 0.8 moles undissociated and 0.2 mole dissociated. Total number of moles of acid =0.1+0.9=1.0 mol Total number of moles of acid =0.2 +0.8=1.0 mol Number of moles of KOH required = 1.0 mole Number of moles of KOH required = 1.0 mole