Asked by Veronica
In what ratio must we use acetic acid pka 4.75 and sodium acetate to produce a buffer system with a pH of 3.86?
The answer key states the following, but I do not understand where the 1/7.75 come from:
[NaOAc]/[HOAc]= 0.129 --> [NaOAc]/[HOAc]= 1/7.75
The answer key states the following, but I do not understand where the 1/7.75 come from:
[NaOAc]/[HOAc]= 0.129 --> [NaOAc]/[HOAc]= 1/7.75
Answers
Answered by
Anonymous
Manipulation of the equation:
pH=pka+log[NaOAc]/[HOAc]
pH-pka=log[NaOAc]/[HOAc]
10^(pH-pka)=[NaOAc]/[HOAc]
10^(3.86-4.75)=[NaOAc]/[HOAc]
10^(-0.89)=[NaOAc]/[HOAc]
0.1288=[NaOAc]/[HOAc] or 7.75=[HOAc]/[NaOAc]
1/0.1288=7.75
pH=pka+log[NaOAc]/[HOAc]
pH-pka=log[NaOAc]/[HOAc]
10^(pH-pka)=[NaOAc]/[HOAc]
10^(3.86-4.75)=[NaOAc]/[HOAc]
10^(-0.89)=[NaOAc]/[HOAc]
0.1288=[NaOAc]/[HOAc] or 7.75=[HOAc]/[NaOAc]
1/0.1288=7.75
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