Question

How many grams of dry NH4Cl need to be added to 1.80 L of a 0.400 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.83? Kb for ammonia is 1.8×10−5.

Answer 1

first calculate the pKa value from Kb

Ka x Kb = Kw = 1.0 x 10^-14

Ka = 1.0 x 10^-14 / 1.8×10⁻⁵

Ka = 5.56 x 10^-10

pKa = -log Ka = -log5.56 x 10^-10  = 9.26

now use the handerson equation

pH = pKa + log([salt/base])

8.83 = 9.25 +  log([salt/base]) here salt = NH4Cl and NH3 is the base

log([salt/base]) = 8.83-9.25 = -0.42

[salt/base] = 10^-0.42 = 0.38

[NH3] given = 0.4 M

[salt/0.4] = 0.38

salt = 0.38 x 0.4 = 0.152 M

now Molarity = moles / Volume in liters

0.152 = moles / 1.8L

moles of NH4 = 0.152 mol.L x 1.8 L

= 0.2736 moles

now use the formula

moles = mass / molar mass

mass = moles x molar mass

= 0.2736 moles x 53.5 g/mol

= 14.63 grams