Question

Just as pHpH is the negative logarithm of [H3O+][H3O+], pKapKa is the negative logarithm of KaKa

pKa=−logKapKa=−log⁡Ka

The Henderson-Hasselbalch equation is used to calculate the pHpH of buffer solutions:

pH=pKa+log[base][acid]pH=pKa+log⁡[base][acid]

Notice that the pHpH of a buffer has a value close to the pKapKa of the acid, differing only by the logarithm of the concentration ratio [base]/[acid][base]/[acid]. The Henderson-Hasselbalch equation in terms of pOH and pKbpKb is similar.

pOH=pKb+log[acid][base]

How many grams of dry NH4ClNH4Cl need to be added to 1.80 LL of a 0.100 MM solution of ammonia, NH3NH3, to prepare a buffer solution that has a pHpH of 8.85? KbKb 1.8×10−51.8×10−5.

Express your answer with the appropriate units.

Answer 1

Using eqn 1 and eqn 2

moles of NH₄⁺/0.180 mol = 2.57

moles of NH₄⁺ = 2.57 * 0.180 mol

= 0.463 mol

Now, the source of NH₄⁺ is the dry NH₄Cl which is added.

So, moles of NH₄Cl to be added = 0.463 mol

Mass of NH₄Cl to be added = mol * molar mass

= 0.463 mol * 53.5 g/mol

= 24.8 g

Hence, 24.8 g of NH₄Cl must be added to 1.80 L of 0.100 M NH₃ to form a buffer of pH 8.85