TBR G Chem Section 5 Titration and Buffers example 5.2

[jw1985]

Hi All,

I was hoping to get some help in understanding this question below. I don't get how the one third plays a role, and the two third works.

Example 5.2
Which of the following solutions results in a buffer with a pH of 5.0, given that HA has a pKa of 4.7?
A. HA with one-half equivalent of A"
B. A" with one equivalent of HA
C. HA with one-third equivalent of OH"
D. A" with one-third equivalent of H30+

The answer is D.

Thanks for your help.

---

Members don't see this ad.

 

[FeinMS]

pH>pKA so by HHB eq, [A-]>[HA].
A. [HA]=1 eq. and [A-]=0.5 eq. So [HA]>[A-]. Eliminate
B. [A-]=1 eq. and [HA]=1 eq. So [HA]=[A]. Eliminate
C. [HA] is initially 1 eq, but adding 1/3 equivalent of OH- deprotonates 1/3 eq of HA to form A-. So [HA]=2/3 eq. [A-]=1/3 eq. Still [HA]>[A]. Eliminate
D. Same logic as C. [A-]=2/3 eq. [HA]=1/3 eq. [A-]>[HA] Yes.

 

[Rucap09]

Just came across the same problem.

Doesn't 1/3 equivalent mean an amount equivalent to 1/3 of the BASE? It sounds like the ratio would come out to being (3/4)/(1/4) which would be log 3.

Is equivalent in this sense always used as a fraction of the total?