# General Chem

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A weak acid has Ka= 1*10^-3. If [HA]= 1.00 M what must be [A-] for the pH to be pH 2.7?

My work:
pH= Pka + log(B/A)
2.7= 3 + log (B/A)

Kw= Ka* Kb
1*10^-14= 1*10^-3 * Kb
Kb= 1*10^-11

I don't know how to go about from here

• General Chem -

weak acid has Ka= 1*10^-3. If [HA]= 1.00 M what must be [A-] for the pH to be pH 2.7?

My work:
pH= Pka + log(B/A)
2.7= 3 + log (B/A)
You are OK to here. You need not use any of what follows. The problem gives you (HA)= 1.00 M [the acid in the problem) and it asks you to solve for B (the (A^-) in the problem).
Kw= Ka* Kb
1*10^-14= 1*10^-3 * Kb
Kb= 1*10^-11

• General Chem -

This method gives me .74M as the answer, and the correct answer is .5M

this is what I did:
2.7= 3 + log (B/A)
-0.3= log(B/1.00)
.74= B/1
B= .74M

I don't understand how it could be 0.5M

• General Chem -

the antilog of -.3 is .5, not .74

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