LarryLute
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Failed Exam question: Predicting pH of buffer solutions.
Hi all!
I participated in an exam sometime ago and there was one question (with 5 parts "a, b, c, d, e") that I didn't know how to solve. Unfortunetly I have
not been able to get my hands on solved answer, so after about a year I desided to ask you.
How is this solved?
Every part, "a, b, c, d, e" can only be answered: "1:1 or 10:11 or 9:10 or 1:2 or 1:3 or 1:5 or 1:9 or 1:10 or 1:11".
It is presumed that pKa of acetic acid is 4,74.
The question:
"
What ratio, in volumes, of the following reagents are mixed to give the pH shown?
A) 2M acetic acid and 1M sodium acetate to give pH of 4,74
B) 1M sodium hydroxide and 1M acetic acid to give pH 5,74
C) 1M acetic acid and 1M sodium hydroxide to give pH of 3,74
D) 1M HCl and 1M sodium acetate to give pH of 5,05
E) 1M HCl and 1M sodium acetate to give pH of 3,74
"
Thanks!
[Edited on 9-5-2014 by LarryLute]
[Edited on 9-5-2014 by LarryLute]
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DraconicAcid
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These are farily easy if you're familiar with the H-H equation (Hendersson Hasselbalch, or something like that).
>What ratio, in volumes, of the following reagents are mixed to give the pH shown?
>A) 2M acetic acid and 1M sodium acetate to give pH of 4,74
To get pH = pKa, you want a 1:1 ratio of acetate to acid, so a 1:2 ratio of these solutions.
> B) 1M sodium hydroxide and 1M acetic acid to give pH 5,74
This is one unit above pKa, so you want ten times as much acetate (formed from the neutralization of acid) as acid. 10:11 ratio.
> C) 1M acetic acid and 1M sodium hydroxide to give pH of 3,74
One unit below pKa- you want ten times as much acid as acetate. 11:1
> D) 1M HCl and 1M sodium acetate to give pH of 5,05
This is 0.31 units above the pKa, and 0.31 is the log(2), so we want twice as much acetate as acid. 1:3 ratio.
>E) 1M HCl and 1M sodium acetate to give pH of 3,74
One unit below, so ten times as much acid as acetate, so 10:11.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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LarryLute
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Part A.. you said "To get pH = pKa, you want a 1:1 ratio of acetate to acid, so a 1:2 ratio of these solutions."
But don't both reagents start reacting cos they are mild acid and its conjugate base. So even if they are added in 1:2, do they really end up as a 1:2
ratio?
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blogfast25
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Quote: Originally posted by LarryLute  | Part A.. you said "To get pH = pKa, you want a 1:1 ratio of acetate to acid, so a 1:2 ratio of these solutions."
But don't both reagents start reacting cos they are mild acid and its conjugate base. So even if they are added in 1:2, do they really end up as a 1:2
ratio?
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1 part of 1 mol/L of the acid reacts with 1 part of 1 mol/L of NaOH to form 2 parts of 0.5 mol/ of NaOAC. Leaving parts 2 parts of 0.5 mol/L of HOAc.
The molar ratio HOAc/NaOAc is thus 1 and pH = pKa.
The real concentrations are slightly different but because HOAc is a weak acid and NaOAc a weak base their concentrations are about the same
as calculated.
[Edited on 15-5-2014 by blogfast25]
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