Question

How many grams of dry NH4Cl need to be added to 1.70 L of a 0.700 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

NH4 is the conjugate base for NH3. You will need to calculate pKa for NH4. Then you can use the Henderson Hasselbalch equation to find the concentration ration [base]/[acid]. Once you know the ratio, you can determine the required molar concentration of the salt and then calculate the mass of the salt.

Calculate pKa for NH4

Kb is 1.8x10^-5, where Ka*Kb=Kw
Ka = 1.0x10^-14/(1.8x10^-5)
Ka = 5.5x10^-10
pka = -ln(Ka)
pka = 9.26

Calculate the ration of base to acid using the HH equation and rearrange to have [base]/[acid] by its self.

8.83=9.26 + log[base]/[acid]
[base]/[acid]= 0.37

Since the pH of the buffer is less than pKa that means the buffer should contain more acid than base. The ratio of base to acid is less than 1, which is in agreement with that expectation.

so now calculate the concentration of NH4

[NH3]/[NH4] =0.37
Concentration of NH3 is 0.700M

[NH4]=0.700/0.37
= 1.89 M

Now that you know the CONCENTRATION and have VOLUME, you now solve for moles (re: C= n/V)

moles = 3.21mols

and using molar mass of 53.5g/mol...

mass = 171.735 grams of NH4Cl