Answer
$[HA]_{initial} = 2.26 \times 10^{-3}M$
Work Step by Step
1. Find $[H^+]$
$[H^+] = 10^{-pH} = 10^{-3.26} = 5.5 \times 10^{-4}M$
2. Using the ICE table for the first interaction of the acid with water:
$[HA] = [HA]_{initial} - x$
$[H^+] = [A^-] = x$
- We know that $[H^+] = 5.5 \times 10^{-4}$, so:
$x = 5.5 \times 10^{-4}$
3. Use the Ka equation to find the $[HA]$ in equilibrium:
Ka formic acid$: 1.77 \times 10^{-4}$
$1.77 \times 10^{-4} = \frac{[H^+][A^-]}{[HA]} = \frac{(5.5 \times 10^{-4})^2}{[HA]}$
$[HA] = \frac{3.025 \times 10^{-7}}{1.77 \times 10^{-4}}$
$[HA] = 1.71 \times 10^{-3}$
4. Find the $[HA]_{initial}$
$[HA] = [HA]_{initial} - x$
$1.71 \times 10^{-3} = [HA]_{initial} - 5.5 \times 10^{-4}$
$[HA]_{initial} = 2.26 \times 10^{-3}M$