Prepare a buffer solution

Question

Calculate how you would prepare 1.000 l of a pH 10.25 buffer made from NH3 and NH4Cl. You are provided with 2.000 l of 3.00 M NH3 and 250 g NH4Cl solid. Kb= 1.75x10-5 . Assume that the addition of solid does not result in a volume change.

Answer 1

We can use the Henderson Hasselbalch equation for an alkaline buffer:

pOH = pKb + log [conj.acid] /[ base]

pOH = 14 - pH = 14 - 10.25 = 3.75

The pKb is -log Kb = -log 1.75x10^-5 = 4.76

3.75 = 4.76 + log [NH₄⁺] / [NH₃]

log [NH₄⁺] / [NH₃] = -1.01

[NH₄⁺] / [NH₃] = 0.0977

One approach to making the buffer would be to take 1 L of the 3 M NH₃ solution and add enough NH₄Cl to obtain the desired ratio of 0.0977. Since 1 L of 3 M NH₃ provides 3 mols of NH₃, in order to obtain the desired ratio, we would need x / 3 = 0.0977 and x = 0.293 moles of NH₄Cl. Using a molar mass of NH₄Cl of 53.5 g / mole, we would need 0.293 mols x 53.5 g / mol = 15.7 g of NH₄Cl.

Summary:

weigh out 15.7 g NH₄Cl and add it to the 1.0 L of 3 M NH₃ solution.

Answer 2

Currently from the 2L of 3M NH3 solution, we have this hydrolysis reaction that is in the form of a weak base dissociation:

NH₃ + H₂O <--> NH₄⁺ + OH^-

Let's convert Kb to a pKb

Kb = 1.75x10^-5

pKb = -log(1.75x10^-5) = 4.76

Use the Henderson-Hasselbalch equation for weak bases to determine the relative ratio of [NH₃] and [NH₄⁺] at pH 10.25

First, convert pH to pOH

pH + pOH = 14

pOH = 14-pH = 14-10.25 = 3.75

general form

pOH = pKb + log([conjugate_acid]/[base])

with our species

pOH = pKb + log([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃])= pOH - pKb

[NH₄⁺]/[NH₃] = 10^(pOH-pKb)

[NH₄⁺]/[NH₃] = 10^{(3.75 - 4.76)} = 10^-1.01

[NH₄⁺]/[NH₃]= 0.098

[NH₄⁺] = 0.098*[NH₃] or [NH₃] = 10.2 [NH₄⁺]

We begin with 1L of 3M NH₃ from the 2L solution since we are told the final buffer must be 1L volume.

If we take 1L of 3M NH₃, solution, we have 3 moles NH₃ (1L * 3mol/L = 3mol). We then use the ratio determined earlier to calculate how much of the conjugate acid, NH₄⁺ we are going to need:

[NH₄⁺] = 0.098 * (1L * 3moles/L NH₃) = 0.294 moles of NH4+

Since this will be a 1L fixed volume buffer, and the problem says we are to assume adding solid won't change the volume, these concentration ratios are going to be equivalent to mole ratios in this case.

NH₄⁺ comes from NH₄Cl salt to take from the solid and dissolve in our 1L solution.

We calculate the mass of NH₄Cl (in grams) we need to dissolve to get 0.294 moles of NH₄

Molar Mass NH₄Cl = 53.5g/mol

0.294 moles NH₄⁺ * 1mol NH₄Cl / 1mol NH₄⁺ * 53.5g NH₄Cl / 1 mol NH₄Cl

= 15.7g of NH₄Cl solid

Thus, to prepare our solution of 1L NH₃:NH₄⁺ buffer at pH = 10.25, we dissolve 15.7g of NH₄Cl in 1L of the 3M NH₃ solution provided.

We can check our work backwards, this time using the acid form of the Henderson-Hasselbalch equation

First, obtain pKa from pKb

Kb * Ka = Kw = 1x10^-14

Ka = Kw/Kb = 10^-14 / 1.75 x 10^-5 = 5.71 x 10^-10

pKa = -log(5.71 x 10^-10 ) = 9.24

Weak acid version of the dissociation:

NH₄⁺ + H₂O <--> NH₃ + H₃O⁺

Henderson-Hasselbalch equation for weak acids

pH = pKa + log([conjugate_base]/[acid])

[conjugate_base] = [NH₃] = 3M

[acid] = [NH₄⁺] = 0.294 moles /1L = 0.294M

pH = 9.24 + log (3M/0.294M) = 10.25

We confirm that our preparation makes the buffer with the correct pH.