Question

A 1.0 M solution of a compound with 2 ionizable groups (pKaʼs = 6.2 and 9.5; 100 mL total) has a pH of 6.8. If a biochemist adds 60 mL of 1.0 M HCl to this solution, the solution will change to which pH?

Answer 1

Let the two ionizable groups be HA and A-

we know that

pH = pKa + log[A-/HA]

so

6.8 = 6.2 + log [0.1 - HA] /[HA]

0.6 = log (0.1 - HA) / HA

3.98 = (0.1-HA) / HA

solving we get

HA = 0.02 mol

now

we know that

moles = moalrity x volume (L)

so

moles of A- + moles of HA = 1 x 0.1 = 0.1

so

HA + A = 0.1

so

0.02 + A = 0.1

A = 0.08 mol

now

moles of HCl added = molarity x volume (L)

so

moles of HCl added = 1 x 60 x 10-3

moles of HCl added = 0.06

now

the reaction becomes

A- + H+ ---> HA

so

moles of A- reacted = moles of H+ added = 0.06

so

moles of A- remaning = 0.08 -0.06

so

moles of A- remaining = 0.02

now

moles of HA formed = moles of A- reacted = 0.06

final moles of HA = 0.02 + 0.06 = 0.08

now

pH = pKa + log [A-/HA]

so

pH = 6.2 + log [0.02/0.08]

pH = 5.6

so

the new pH of the solution is 5.6