Question

a buffer solution is prepared by mixing 100cm3 aqueous NH3 0.1mol dm-3 with 100cm3 NH4Cl mol dm-3 . Given that KB for NH3 =1.74x10^-5
...
1) calculate pH of the buffer solution.
...
2)calculate the solution after addition 50cm3 HCl of 0.1 mol dm-3.

Answer 1

KB for NH3 =1.74x10^-5
pKB = 4.76

100cm3 = 0.1 L NH3

0.1mol dm-3 = 0.1 mol / L or 0.1 M

100cm3 = 0.1 L NH4Cl 1.0 mol dm-3 or 1.0 M

NH3 + H2O <------< NH4+ + OH-

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

pOH = 4.76 + log 1.0/ 0.1

=4.76+1.0

=5.76
pH = 14 – 5.76

=8.24

Now you add 0.05 L of 0.100M HCl to the 200mL of buffer.
HCl reacts with NH3 in 1:1 molar ratio to produce NH4Cl
HCl + NH3 → NH4Cl.

Mol HCl = 0.05*0.100 = 0.005 mol HCl
This will react with the NH3 in solution to produce 0.005 mol NH4Cl and the mol of NH3 will be reduced by 0.005 mol

Mol NH3 in 100mL of 0.100M solution = 100/1000*0.1 = 0.01 mol NH3
Mol NH4Cl in 100mL of 0.1M solution = 100/1000*0.1 = 0.01 mol NH4Cl

After adding the HCl and reaction to produce 0.005 mol NH4Cl , you have
Mol NH3 = 0.01-0.005 = 0.005 mol NH3
Mol NH4Cl 0.01+0.005 = 0.015 mol NH4Cl .

then
pOH = pKa + log ([salt] / [base])
pOH = 4.76 + log (0.015/0.005)
pOH = 4.76 + log 3.0
pOH = 4.76 + 0.477
pOH = 5.237

pH = 14.00 – 5.237
pH = 8.763