Question

A 11mM solution of a compound with a pKa=6.2 (1L total) has a pH of 5.2. If a biochemist add 9mmol of NaOH to this solution, the solution will change to a pH:

A) 5.85

B)6.2

C)7.2

D)6.85

E)5.55

Answer 1

Calculate m mol of compound.

m moles of compound = 11 m mol/ L x 1.0 L = 11 m moles

moles of compound remained after reacting with NaOH

= 11 – 9 = 2 m mol

So the concentration would be 2 mM

Concentration in terms of M = 2 m mol / L x ( 1 M / 1000 m M)

= 0.002 M

Same time there is a formation of 9 m mol of NaA

Lets show the reaction

HA + NaOH -- > Na A + H2O

Here Na is also formed and m moles of it = m moles of limiting reactant. Limiting reactant is NaOH

Therefore m moles of Na A = 9 m mol

Lets show it concentration in M

= 9 m M x 1 M / 1000 mM

= 0.009 M

Now use Henderson Hasselbalch equation to find pH

pH = pka + log ( [NaA]/[HA])

lets plug in the values

pH = 6.2 + log ( 0.002 / 0.009)

= 5.55

So the pH of the resultant solution would be 5.5