The acid-dissociation constant at 25.0 °C for hypochlorous acid (HClO)is 3.0 × $10 ^ { - 8 }$ .At equilibrium,the molarity of $\mathrm { H } _ { 3 } \mathrm { O } ^ { + }$ in a 0.030 mol L^-1 solution of HClO is ________. A)1.52 mol L^-1 B)3.3 × 10^-10mol L^-1 C)4.52 mol L^-1 D)0.030 mol L^-1 E)3.0 × 10^-5mol L^-1

Question

Quizplus · Accepted Answer

The equation for the dissociation of HClO is:

$\mathrm { HClO + H_2O ightleftharpoons H_3O^+ + ClO^-}$

The acid dissociation constant expression for this equation is:

$K_a = \frac {\mathrm {[H_3O^+][ClO^-]}} {\mathrm {[HClO]}}$

We can assume that the concentration of $\mathrm {ClO^-}$ is negligible compared to the initial concentration of $\mathrm {[HClO]}$ since its value would be much smaller than the dissociation quotient ($Q_c$) . Therefore, we can simplify the expression as follows:

$K_a = \frac {\mathrm {[H_3O^+][ClO^-]}} {\mathrm {[HClO]}} \approx \frac {\mathrm {[H_3O^+]^2}} {\mathrm {[HClO]}}$

We can rearrange this expression and solve for $\mathrm {[H_3O^+]}$ :

$\mathrm {[H_3O^+]} = \sqrt {\frac {K_a [\mathrm {HClO}]} {[\mathrm {HClO}]}}= \sqrt {K_a}$

Plugging the given value of $K_a$ into this equation gives:

$\mathrm {[H_3O^+]} = \sqrt {3 	imes 10^{-8}} = 5.48 	imes 10^{-5} \mathrm{mol/L}$

This value represents the equilibrium concentration of $\mathrm {H_3O^+}$ in the solution. Since the initial concentration of HClO is given as 0.030 M, and we assume that most of it dissociates, we can say that the approximate concentration of $\mathrm {H_3O^+}$ is also 0.030 M. Since the equilibrium concentration is much smaller than the initial concentration, we can assume that the approximation is valid. Therefore, the answer is:

$\mathrm {[H_3O^+]} = 5.48 	imes 10^{-5} \mathrm{mol/L} \approx 3 	imes 10^{-5} \mathrm{mol/L}$

The answer is closest to option E, 3.0 × 10^-5 mol/L, which is within the expected range of values.

The Acid-Dissociation Constant at 25 10−810 ^ { - 8 }10−8

The Acid-Dissociation Constant at 25 $10 ^ { - 8 }$