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
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