Question

\\\frac{(0.1402 + x)x}{0.1402 - x} = 1.549 \times 10^{-4}\\

lets try the weak acid approximation:

\\\frac{(0.1402 + x)x}{0.1402 - x} \approx \frac{0.1402x}{0.1402} = 1.549 \times 10^{-4}\\

\\x = 1.549 \times 10^{-4}\\

check the approximation:

\\\frac{1.549 \times 10^{-4}}{0.1402} = 0.001105... = 0.11\\%\\ looks good.

since \\(x\\) is equal to the equilibrium molarity of \\(\text{c}_2\text{o}_4^{2-}\\), you have the result you need. be sure to round your final answer to 2 significant figures.

correct answer

\\1.5 \times 10^{-4}\text{ m}\\

Explanation:

Resolver la ecuación de equilibrio químico

$$ \frac{(0.1402 + x)x}{0.1402 - x} = 1.549 \times 10^{-4} $$

Aplicar la aproximación de ácido débil

$$
LATEXBLOCK0
$$

Redondear a dos cifras significativas con unidades

$$ x \approx 1.5 \times 10^{-4}\text{ M} $$

Answer:

\(1.5 \times 10^{-4}\text{ M}\)