Question

How many grams of dry NH4Cl need to be added to 2.20 L of a 0.600 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.50? Kb for ammonia is 1.8×10−5.

Answer 1

Solution :-

Formula to calculate the pH of the buffer solution is

pOH = pkb + log [acid /base]

pOH = 14 - pH

pOH = 14 - 8.50 = 5.5

pkb = -log kb

pkb = - log 1.8*10^-5

pkb =4.74

now lets use the values in the formula and calculate the molarity of the conjugate acid

pOH = pkb + log [acid /base]

5.50 = 4.74 + log [acid / 0.600]

5.50 - 4.74 = log [acid / 0.600]

0.76 = log [acid / 0.600]
antilog 0.76 = [acid / 0.600]
5.75 = acid / 0.600

5.75 * 0.600 M = [acid]

3.45 M = [acid]

Now lets calculate the moles of NH4Cl using the molarity and volume

moles = molarity * volume

moles of NH4Cl = 3.45 mol per L * 2.20 L = 7.59 mol NH4Cl

now lets convert moles of NH4Cl to its mass

mass= moles * molar mass

mass of NH4Cl = 7.59 mol * 53.491 g per mol

                      = 406 g NH4Cl

So we need to add 406 g of dry NH4Cl to get the pH 8.50