In: Chemistry
Which of the following acids would be most suitable for preparing a buffer of pH 3.10 and why: hydrogen peroxide (pKa= 11.65), propanoic acid (pKa=4.874), cyanoacetic acid (pKa= 2.472), and 4-aminobenzenesulfonic acid (pKa=3.232)?
Which of the following acids would be most suitable for preparing a buffer of pH 3.10 and why: hydrogen peroxide (pKa= 11.65), propanoic acid (pKa=4.874), cyanoacetic acid (pKa= 2.472), and 4-aminobenzenesulfonic acid (pKa=3.232)?
We want to prepare a buffer with pH = 3.10 i.e. an acidic buffer.
Buffer is formed from the mixing of acid and it salt solution with particular concentration ratio i.e. [Acid]:[Salt] ratio.
The relation between pH of buffer, pKa of acid used, and the [Acid]:[Salt] ratio are related by equation well known as Henderson equation,
pH = pKa + log{[Acid]/[Salt]}
We are given with 4 acids with their pKa values. For all of them using Henderson equation let us calculate [Acid]:[Salt] ratio.
1)H2O2(HOOH), pKa = 11.65,
pH = pKa + log{ [Acid]/ [Salt]}
3.10 = 11.65 + log{ [HOOH]/[‑OOH]}
3.10 - 11.65 = log{ [HOOH]/[‑OOH]}
log{ [HOOH]/[‑OOH]} = -8.55
[HOOH]/[‑OOH] = Antilog(-8.55)
[HOOH]/[‑OOH] = 2.818 x 10-9.
2)Propanoic acid, pka =4.874
log{[Propanoic acid]/[Propanoate]} = pH – pKa = 3.10 – 4.874 = -1.774
[Propanoic acid]/[Propanoate] = Antilog(-1.774)
[Propanoic acid]/[Propanoate] = 0.0168
3)Cyanoacetic acid, pKa = 2.472
Log{[Cyanoacetic acid]/[Cyanoacetate]} = 3.10-2.472 = 0.628
[Cyanoacetic acid]/[Cyanoacetate] = Antiolog(0.628)
[Cyanoacetic acid]/[Cyanoacetate] = 4.42
4)4-Aminibenzenesulfonic acid(ABSA), pKa =3.232
Log{[ABSA]/[Salt]} = pH-pKa = 3.10-3.232 = -0.132
[ABSA]/[Salt] =Antilog(-0.132) = 1.36
i.e. [ABSA] = 1.36[its salt}
Thus, ratio is simplest for 4-Aminibenzenesulfonic acid and hence most suitable for preparing buffer with pH 3.10.
Note: It is always better to use such an acid whose pKa is in desired pH range.
If we use equimolar acid and salt solution we need to take them with just volume ratio as above ratios
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