How many moles of NH4Cl must be added to 2.0L of a 0.10M NH3 to form a buffer whose pH is 9.00? (KbNH3 = 1.8*10^-5)
Solution: This is a buffer question. A buffer solution consists of a weak acid and its conjugate base. In this case NH4+ is the weak acid (comes from NH4Cl) and NH3 is its conjugate base.
Whenever the question states that we have a buffer, we can use Henderson Hasselbach equation.
pH = pKa + log [A-]
[HA]
In this equation pKa = -log (Ka)
[A-] = weak conjugate base
[HA] = weak acid
We are given Kb, and can find Ka. Ka*Kb = 1*10^14
Ka = 5.6*10^-10
Now, we can plug in the values: 9.00 = -log( 5.6*10^-10) +log [A-]
[HA]
log [A-]. = -.2518
[HA]
[A-] = .56
[HA]
[A-] = 0.10 according to the given (molarity of NH3)
0.10 = .56
[HA]
[HA] = 0.1786M
Molarity = moles/ L
0.179M = moles/ 2.0L
moles = .357