Question

How many grams of NaH4Cl need to be added to 1.80L of a 0.0500M solution of ammonia NH4, to prepare a buffer solution that has a pH of 8.76? Kb for ammonia is 1.8*10^5​- ​. Any help would be great!

Answer 1

pH = pKa + log([A^-]/[HA])

Here, [A^-] = [NH3] (base)
[HA] = [NH4^+] (acid)

Rather than use Kb, it is easier to use the pKa of NH4^+ (the conjugate acid).

Kb = 1.8 x 10^-5
pKb = -log(Kb) = -log(1.8 x 10^-5) = 4.74

pKa + pKb = 14
pKa = 14 - pKb = 14 - 4.74 = 9.26

pH = pKa + log([A^-]/[HA])
8.76 = 9.26 + log((0.500 M NH3)/(x M NH4^+))
-0.5 = log((0.500 M NH3)/(x M NH4^+))
(0.500 M NH3)/(x M NH4^+) = 10^-0.5 = 0. 0.3162
x M NH4^+ = (0.500 M NH3 )/(0.3162)
x M NH4^+ = 1.58 M NH4^+

So we need the NH4CL concentration to be 1.58 M. We have 1.8 L of solution, so the number of moles of NH4Cl we need are

(1.58 M NH4Cl)*(1.80 L) = 2.846 moles of NH4Cl

The mass of this number of moles is then

(2.846 moles of NH4Cl)*(53.491 g/mol) = 152.25 g of NH4Cl.