Posted by **Hina** on Sunday, August 9, 2009 at 10:02pm.

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 -
**DrBob222**, Sunday, August 9, 2009 at 10:17pm
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 -
**Anonymous**, Monday, August 10, 2009 at 12:26am
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 -
**bobpursley**, Monday, August 10, 2009 at 4:29am
the antilog of -.3 is .5, not .74

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