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
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
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
the antilog of -.3 is .5, not .74
To find the concentration of A- ([A-]), you can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
You have already calculated the pKa = 3 (as Ka = 1 x 10^-3), and you want to find [A-] given that the pH is 2.7 and [HA] = 1.00 M.
Substituting the known values:
2.7 = 3 + log([A-]/1.00)
Rearranging the equation:
log([A-]/1.00) = 2.7 - 3
log([A-]/1.00) = -0.3
Taking the antilog (reversing the log function) of both sides:
[A-]/1.00 = 10^(-0.3)
Simplifying further:
[A-] = 1.00 x 10^(-0.3)
Calculating this value:
[A-] = 0.501 M (rounded to three decimal places)
Therefore, the concentration of A- ([A-]) must be approximately 0.501 M for the pH to be 2.7 when [HA] = 1.00 M.