Question

We wish to prepare a buffer of pH 10 using common laboratory reagents. How many mL of 6M HCl would we have to add to a solution prepared by adding 100mL of 15M NH3 to 200mL of water?

Ka of NH4+ = 5.7 x 10^-10

pka = 9.24

Please explain how you found the values for the table.

Answer 1

100 mL of 15M NH3 is diluted by adding 200 mL of water. Thus the total volume of NH3 solution is 300mL.Its molarity is [100x15]/300 = 5M

Let us assume VmL of HCl to be added to make the buffer.

NH₃   + HCl -----> NH₄Cl

300 x5 0 0 initial mmoles

1500 6V 0 Change

1500-6V 0 6V after equilibrium

Given pKa of NH4+ = 9.24, hence pKb of NH3 is [14-9.24] = 4.76

The pH of the required buffer=10 and thus pOH of the buffer = 14-10 = 4.0

The pOH of the basic buffer is given by Hendersen equation as

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

4.0 = 4.76 + log [6V/ (1500-6V)]

solving for V we get V = 37.016mL

Thus the volume of 6M Hcl required is 37.016mL

Substituting the given values