The first thing to note here is that you need the value of ammonia's base dissociation constant,
Kb, which is listed as being equal to Kb =1.8 x 10⁻⁵
Now, you can solve this relatively quickly by using the Henderson - Hasselbalch equation to find the pOH of a buffer solution that contains ammonia, a weak base, and the ammonium ion, its conjugate acid
pOH = pKb + log([conjugate acid][weak base])
Start by calculating the pOH of the target solution
pOH=14 - pH
pOH=14 - 9.55 = 4.45
Use the base dissocation constant to find pKb
pKb=-log([OH-])
pKb=-log(1.8 x 10⁻⁵)=4.74
Plug these values into the Henderson - Hasselbalch equation to get
4.45 = 4.7 + log([NH+4][NH3])
log([NH+4][NH3])=4.45 - 4.74
This is equivalent to
[NH+4][NH3]=10^0.29
[NH+4][NH3]=0.5129
This means that the ratio between the concentration of the conjugate acid, which in your case will be provided by the ammonium chloride, and the concentration of the base must be equal to
0.7483
.
The concentration of the conjugate acid will thus be 0.08206 M
This means that you will need to provide the solution with
c=n
Moles = concentration * volume
Since ammonium chloride dissociates in aqueous solution to give
NH₄Cl(aq)↔ NH⁺⁴(aq)+Cl⁻(aq)
it follows that the number of moles of ammonium chloride will be equal to the number of moles of ammonium ions.
nNH₄Cl = nNH₄=0.2175 moles
To find how many grams of ammonium chloride would contain this many moles, use the compound's molar mass
0.2175 moles * 53.49 g / mole=11.634 g NH₄Cl
Rounded to three sig figs, the answer will be mNH₄Cl=11.6 g
