Question

How much 5M KOH must be added to a 1.0L of 0.1M glycine at pH 9.0 to bring its pH to exactly 10.0? When 99% of the glycine is in its protonated form, what is the numerical relation between the pH of the solution and the pKA of the amino group?

Answer 1

We would use Hendersen-Hasselbalck equation,

pH = pKa + log[A-]/[HA]

1 mole of KOH would result in 1 mole of [A-] salt formation

we have from above data,

moles of glycine = M x L = 0.1 x 1 = 0.1 moles

pKa of glycine-NH2 is 9.6

For pH 9.0

9.0 = 9.6 + log[A-]/[HA]

or,

[A-] = 0.25[HA]

we know, equal volume of [base]+[acid] = 0.1

Feed value for [A-] from above

0.25[acid] + [acid] = 0.1

1.25[acid] = 0.1

[acid] = 0.08 mol, so [base] = 0.02 mol

For pH 10.0

10.0 = 9.6 + log[A-]/[HA]

or,

[A-] = 2.512[HA]

again feeding [A-] from above,

2.5[acid] + [acid] = 0.1

[acid] = 0.03 mol, [base] = 0.07 mol

the diference between the two moles of [base] is 0.05 mol will be the amount of base required to be added

So, volume of KOH = moles/M = 0.05/5 = 0.010 L = 10 mL

thus the volume of KOH required to add would be 10 mL to go from pH 9 to pH 10

Now, when 99% of glycine is deprotonated, that is [A-] = 0.99 and [HA] = 0.01

we will have,

pH = pKa + log(0.99/0.01)

     = pKa + 2.00

So, pH of solution would be pKa + 2.00

If we substitute pKa = 9.6,

pH = 11.6 will be for this resulting solution