What mass of ammonium chloride should be added to 2.45 l of a 0.165 m nh3 in order to obtain a buffer with a ph of 9.50?

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

NH ₄ Cl (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.

nNH ₄ Cl = nNH ₄ = 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 NH ₄ Cl

Rounded to three sig figs, the answer will be mNH ₄ Cl=11.6 g