# CH_2H_3O_2(aq) + H_2O(l) rightleftharpoons H_3O^+ + C_2H_3O_2^-(aq). K_c = 1.8 X 10^-5 at 25 "^oC. If a solution initially contains 0.210 M HC_2H_3O_2, what is the equilibrium concentration of H_3O^+ at 25 "^oC?

Oct 15, 2016

$\textsf{\left[{H}_{3} {O}^{+}\right] = 1.9 \times {10}^{- 3} \textcolor{w h i t e}{x} \text{mol/l}}$

#### Explanation:

Set up an ICE table based on concentrations in mol/l:

$\textsf{\text{ } C {H}_{2} {H}_{3} {O}_{2} + {H}_{2} O r i g h t \le f t h a r p \infty n s {C}_{2} {H}_{3} {O}_{2}^{-} + {H}_{3} {O}^{+}}$

$\textsf{\textcolor{red}{I} \text{ "0.210" "0" } 0}$

$\textsf{\textcolor{red}{C} \text{ "-x" "+x" } + x}$

$\textsf{\textcolor{red}{E} \text{ "(0.210-x)" "x" } x}$

Applying the equilibrium law:

$\textsf{{K}_{c} = \frac{{x}^{2}}{\left(0.210 - x\right)} = 1.8 \times {10}^{- 5}}$

Because the dissociation is so small we can make the approximation:

$\textsf{\left(0.210 - x\right) \Rightarrow 0.210}$

$\therefore$$\textsf{{x}^{2} = 0.210 \times 1.8 \times {10}^{- 5} = 0.378 \times {10}^{- 5}}$

$\therefore$$\textsf{x = \sqrt{0.378 \times {10}^{- 5}} = 1.9 \times {10}^{- 3} \textcolor{w h i t e}{x} \text{mol/l}}$