Answer:
Amount of NH4Cl that must be added = 61.5 g
Explanation:
Given:
Concentration of ammonia, [NH3] = 0.500 M
Volume of solution = 2.00 L
pH of buffer = 9.20
Kb (NH3) = 1.80*10^-5
To determine:
The amount of NH4Cl to be added
Explanation:
NH3/NH4Cl is a basic buffer, the corresponding Henderson-Hasselbalch equation is:
[tex]pOH = pKb + log\frac{[BH+]}{[B]}[/tex]
where kb = base dissociation constant
BH+ = conjugate acid = NH4Cl
B = base = NH3
It is given that pH = 9.20, therefore,
[tex]pOH = 14-pH = 14-9.20 = 4.8[/tex]
[tex]pKb = -logKb = -log(1.80*10^{-5})=4.74[/tex]
Substituting the appropriate values into the Henderson equation gives:
[tex]4.8 = 4.74 + log\frac{[NH4Cl]}{[0.500]}[/tex]
[NH4Cl] = 0.574 M
[tex]Moles(NH4Cl)=Molarity*Volume = 0.574moles/L*2.00L=1.15 moles[/tex]
[tex]Mass(NH4Cl)=Moles*Mol.wt = 1.15*53.49 = 61.5 g[/tex]
