You dispose of a $0.200\;M$ solution of $H_3C_6H_5O_7$
(citric acid, $K_a$ $=$ $3.5\cdot 10^{-4}$)
Equilibrium of hydrolysis:
$H_3C_6H_5O_7$ $+$ $H_2O$
$H_3O^+$ $+$ $H_2C_6H_5O_7^-$
Substitute now the equilibrium concentration in $K_a$ ! Which of the following substitutions is correct?
a) $3.5 \cdot 10^{-4}$ $ = $ $\frac{x(0.200-x)}{ x^2}$ b) $3.5 \cdot 10^{-4}$ $ =$ $ \frac{0.200-x }{ x^2}$ c) $3.5 \cdot 10^{-4}$ $ =$ $ \frac{x^2 }{ 0.200-x}$ d) $3.5 \cdot 10^{-4}$ $ =$ $ \frac{x^2 }{ x(0.200-x)}$
The answer c) is correct, $3.5 \cdot 10^{-4}$ $ =$ $\frac{x^2 }{ 0.200-x}$