Question

You have a 750mL of a buffer solution that is composed of NH3 and NH4Cl with a Kb value 1.8x 10^5 and a pH 9.43...... How many mL of 6.00M NaOH can be added before the buffer is exhausted?

Answer 1

The pH of buffer is given byHnedersen equation as

pOH = pKb + log [conjugate acid]/[base]

and pH = 14 -pOH

Given pH of buffer = 9.43

pOh = 14-9.43 = 4.57

and pKb of base = 5-log 1.8 = 4.74

Thus

pOH = pKb +log [NH4+]/[NH3]

4.57 = 4.74 + log[NH4+]/[NH3]

Thus [NH4+]/[NH3] =  0.676

Thus [NH4+] is 0.676 times more than NH3 .or if [NH3] = 1 M, [NH4+] = 0.676 M

By the addition of NaOH the NH4+ only reacts. So to exhuast all the NH4+

NH4+ + NaOH -----------------> NH3 + H2O + Na+

0.676 x750 0 1.0x750

=507 = 750 initial mmoles

Thus mmoles of NaOH = V x 6 = 507

Thus V = 84.5 mL

Thus the volume of 6M NaOh required to exhaust the given buffer = 84.5 mL (assuming [NH3] =1M and [NH4+] =0.676M as no total concentration ws given]