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The ph of a 0.10 Msolution of formic acid, HCOOH, at 25 c is 2.38. Calculate Ka for formic acid at this temperature. we know that
Accepted Answer
To calculate Ka for formic acid (HCOOH), we can use the pH and the initial concentration of the acid. Here's how:
1. Write the equilibrium expression:
HCOOH(aq) ⇌ H+(aq) + HCOO-(aq)
2. Set up an ICE table (Initial, Change, Equilibrium):
| | HCOOH | H+ | HCOO- |
|---------|--------|-------|--------|
| Initial | 0.10 M | 0 | 0 |
| Change | -x | +x | +x |
| Equil. | 0.10-x | x | x |
3. Relate pH to [H+] and Ka:
pH = -log[H+] => [H+] = 10^-pH = 10^-2.38 = 4.17 x 10^-3 M
Ka = [H+][HCOO-] / [HCOOH] = (x)(x) / (0.10 - x)
4. Substitute and solve for Ka:
Ka = (4.17 x 10^-3)^2 / (0.10 - 4.17 x 10^-3) = 1.8 x 10^-4
Therefore, the Ka for formic acid at 25°C is approximately 1.8 x 10^-4.
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