Take Home Test Aqueous equilibrium What combination of a molecular base (i.e a n
ID: 993190 • Letter: T
Question
Take Home Test Aqueous equilibrium What combination of a molecular base (i.e a neutral compound not an ion!) and a salt of conjugate acid of that molecular base should you select to a prepare a buffer solution that would maintain a pH of approximately 9.50? explain your choice (in a simple, straightforward manner) What ratio of concentrations of the conjugate acid and molecular base you selected in of this problem would be necessary to achieve a pH of exactly 9.50? 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 prime m looking for volume, concentration and grams.Explanation / Answer
I will answer part a) and part b) for now, part c) post it in another question thread:
a) for a pH of 9.50 of a basic made of bases, means that the pOH would have to be:
pOH = 14-9.5 = 4.5
so, in order to achieve a buffer with this conditions we need a base with a pKb closer to this value of pOH. Looking at the literature we can say that one we can use is the CN- and HCN (pKb = 4.65) or even NH3/NH4Cl (pKb = 4.75), these two could made a great buffer. The more correct would be the CN- but the problem state that it should be a molecular base and a salt of conjugate acid so, I will use this the buffer NH3/NH4Cl (You can also use the CN-/HCN buffer)
b) the ratio to get the exact 9.50 pH would be: (I'll call NH4+/NH3 as R for ratio)
4.5 = 4.75 + log(NH4+/NH3)
4.5-4.75 = logR
10-0.25 = R
R = (NH4+/NH3) = 0.5623
If we use the HCN/CN the result would be:
4.5 = 4.69 + logR
10-0.19 = R
R = 0.645
if lower in the ammonia, so this would be better buffer.
Hope this helps
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