Answer:
Weak acid
Explanation:
The key to this question is understanding what we mean by
pH and pKa, which are logarithmic functions. Students tend to have problems with the logarithmic function. I will introduce the subject briefly.
When I write logab =c, I am asking to what power I raise the base a, to get b. Here ac = b. Now, usually we use the bases 10 or e. So log10100 = 2, log101000 = 3, log101000000 = 6. Likewise log100.1 = log1010 − 1 = −1. In the days before electronic calculators (approx. 30-40 years), students would be issued log tables so that complicated calculations could be performed.
From the above
pH = −log10[H3O +], and pKa = −log10Ka. These are simple functions that have been widely used in chemistry.
Now it is fact, that when a weak acid is mixed with its conjugate base in appreciable concentrations, a buffer solution is formed that tends to resist gross changes in
pH.
We can write, pH = pKa + log10 {[A−][HA]}.
It is clear that when [HA] = [A−], then pH = pKa because log10 {[A−][HA]} = log101 = 0
So we want an acid whose pKa ≅ pH . To get pKa
I simply perform the function
−log10Ka, on each of the acid dissociation constants.
pKaHA = 2.57; pKaHB = 5.36; pKaHC = 8.58;
We want to maintain pH at 5.5, so it is clear that acid HB
is the acid of choice
= 28 + (2 x 16) = 60 g
= 2(18) = 36 g