Please help! There are three parts to this. A. What combination of a molecular b
ID: 992723 • Letter: P
Question
Please help! There are three parts to this.
A. What combination of a molecular base (i.e. a neutral compound not an ion!) and a salt of the conjugate acid of that molecular base should you select to prepare a buffer solution that would maintain a pH of approximately 9.50?
B. What ratio of concentrations of the conjugate acid and molecular base you selected in part a of this problem would be necessary to achieve a pH of exactly 9.50?
C. Describe, using one short well-written sentence, a simple and practical way to make 1.00L of the buffer solution you have defined in parts a and b of this problem. Assume that a 1.0 M solution of the molecular base is available and that the chloride salt of the conjugate acid is available as a pure solid. Note: Even if you have not had a lab course you should be able to provide more details than the number of moles to combine, I’m looking for volume, concentration and grams.
Explanation / Answer
I believe you are talking about ammonia buffer solution. The salt of the conjugate acid of ammonia is ammonium chloride, which is easily available (note the problem mentions that the chloride salt of the conjugate acid is available as a solid). Hence, my proposed answers are:
(a) The molecular base that is used is aqueous ammonia (NH4OH) (pka = 9.25) and the salt of the conjugate acid that is taken is ammonium chloride (NH4Cl).
(b) The pH of the buffer solution is 9.50 and the pKa of aqueous ammonia is 9.25. We shall consider the dissociation of ammonium ion, since we know the pKa value of the conjugate acid of ammonia.
NH4+ (aq) + H2O (l) --------> H3O+ (aq) + NH3 (aq)
Ka = [H3O+][NH3]/[NH4+]
Taking logarithm, we have,
log10Ka = log10{[H3O+][NH3]/[NH4+]}
or, log10Ka = log10[H3O+] + log10[ NH3]/[NH4+]
Taking negative on both sides, we have
- log10Ka = -log10[H3O+] – log10[NH3]/[NH4+]
pKa = pH – log10[NH3]/[NH4+] (these two terms are the pKa and pH)
or, pH = pKa + log10[NH3]/[NH4+]
Putting the values, we have,
9.50 = 9.25 + log10[NH3]/[NH4+]
===> 0.5 = log10[NH3]/[NH4+]
or, [NH3]/[NH4+] = 100.5 = 3.16
Therefore, [NH3]/[NH4+] = 3.16/1
A molar ratio of NH3: NH4+ (base: conjugate acid) in 3.16: 1 is required to prepare the buffer of pH 9.50
(c) We can prepare the buffer solution by taking a 1.00 L of 1.0 M ammonia solution.
Molar mass of NH4Cl is 53.5 gm/mol.
If we dissolve 53.5 gm of the ammonium chloride in this 1.00 L solution, the molar concentration will be 1.0 M with respect to NH4Cl.
But as per the molar ratio, the [NH4Cl] = [NH3]/3.16 = (1.00 M)/3.16 = 0.316 M.
To produce 1.00 L 1.0 M solution, we need 53.5 gm NH4Cl.
Hence, the weight of NH4Cl to be taken is (53.5 gm)*(0.316) = 16.93 gm
Hence, we take 1.0 L 1.0 M ammonia solution and dissolve 16.93 gm of solid ammonium chloride into the solution to prepare the buffer of pH 9.50