Question

Discuss how to prepare 250.0 mL of an ammonium–ammonia buffer (pKb = 4.70, pH = 9...

Discuss how to prepare 250.0 mL of an ammonium–ammonia buffer (pKb = 4.70, pH = 9 and total concentration of 0.05 M) using 0.10 M ammonium chloride and 0.10 M aqueous ammonia.

pH = 9 then pOH = 14-pH = 14.9 = 5

so

the buffer equation, from henderson hasselbach

pOH = pKB + loG(NH4+/NH3)

pKB for NH3 = 4.75

5= 4.75 + log(NH4+ /NH3)

total concentration must be 0.05 M them

[NH4+]/[NH3] = 10^(5-4.75) = 1.77827

[NH4+]/[NH3] =1.77827

NH4+/NH3 =1.77827

V = 250 mL = 0.25 L so

total molarity = MV = 250*0.05 = 12.5 mmol of buffer

so:

NH4+/NH3 =1.77827

mmol of NH4+ --> MV = 0.1*(Vconjugate)

mmol of NH3 -- >MV = 0.1*(Vbase)

NH4+/NH3 =1.77827

0.1*(Vconjugate) / 0.1*(Vbase) = 1.77827

Vtotal = 250 mL so

Vconjugate + Vbase = 250

Vbase = 250- Vconjugate

0.1*(Vconjugate) / 0.1*(Vbase) = 1.77827

substitute

0.1*(Vconjugate) / 0.1*(250- Vconjugate ) = 1.77827

1/1.77827*(Vconjugate) = 250- Vconjugate

0.562344*Vconjugate = 250 - Vconjugate

1.56234Vconjugate = 250

Vconjguate = 250/1.56234 = 160.01638 mL

so

Vbase = 250 -160.01638 = 89.9 mL