Answer
$[H^+] = 5.71 \times 10^{-4} M $
$[CH_3COO^-] = 5.71 \times 10^{-4}M$
$[CH_3COOH] = 0.018129M$
Work Step by Step
1. Calculate the initial acid concentration.
- Find the number of moles.
Acetic acid: Molar Mass = 60g/mol.
$mm = \frac{mass}{n(moles)}$
$60 = \frac{0.0560}{n(moles)}$
$n(moles) = \frac{0.056}{60} = 9.333 \times 10^{-4}$
- Find the concentration:
$Concentration(mol/L) = \frac{n(moles)}{V(L)} = \frac{9.333 \times 10^{-4}}{0.050} = 0.0187 M$
2. Draw the ICE table: (Image on the end)
3. Now, use the Ka value and equation to find "x".
$Ka = 1.8 \times 10^{-5} = \frac{(x)*(x)}{0.0187-x} $
$Ka = 1.8 \times 10^{-5} = \frac{(x^2)}{0.0187-x} $
$3.366 \times 10^{-7} - 1.8 \times 10^{-5}x = x^2$
$3.366 \times 10^{-7} - 1.8 \times 10^{-5}x - x^2 = 0$
Bhaskara:
$\Delta = (-1.8 \times 10^{-5})^2 - 4 * (-1) * (3.366 \times 10^{-7})$
$\Delta = 3.24 \times 10^{-10} + 1.346 \times 10^{-6}$
$\Delta \approx 1.346 \times 10^{-6}$
$x_1 = \frac{-(-1.8 \times 10^{-5}) + \sqrt {1.346 \times 10^{-6}}}{2*(-1)}$
or
$x_2 = \frac{-(-1.8 \times 10^{-5}) - \sqrt {1.346 \times 10^{-6}}}{2*(-1)}$
$x_1 = -5.89 \times 10^{-4} (Negative)$
$x_2 = 5.71 \times 10^{-4} $
The concentration can't be negative.
$[H^+] = [A^-] = 0 + x = x = 5.71 \times 10^{-4}$
$[HA] = 0.0187 - x = 0.0187 - 5.71 \times 10^{-4} = 0.018129$
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