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How many moles of NH4Cl must be added to 2.0L of a 0.10M NH3 to form a buffer whose pH is 9? (KbNH3

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

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