Question

How many grams of dry NH4Cl need to be added to 2.00 L of a 0.100 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.53? Kb for ammonia is 1.8×10−5.

Answer 1

Given that;

Kb for ammonia is 1.8×10−5.

pH = 8.53

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

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

first calculate the value of pKb as follows:

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

= 4.74
and we know that;

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

then;

How many grams of dry NH4Cl need to be added to 2.00 L of a 0.100 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.53?

pH = pKa + log([A^-]/[HA])
8.53 = 9.26 + log((0.100 M NH3)/(x M NH4^+))

-0.73 = log((0.400 M NH3)/(x M NH4^+))

(0.400 M NH3)/(x M NH4^+) = 10^-0.73 = 0.186

x M NH4^+ = (0.400 M NH3 )/(0.186)
x M NH4^+ = 2.15 M NH4^+

thus we required the NH4CL concentration to be 2.15 M. here the volume of solution is 2L , so the number of moles of NH4Cl we need are
number of moles = molarity * volume in L

= (2.15 M NH4Cl)*(2 L)

= 4.30 moles of NH4Cl

Amount in g = number of moles * molar mass

= 4.30 moles of NH4Cl * (53.491 g/mol)

= 230 g of NH4Cl