Question

A solution is made initially with 0.200 M HIO3 (Kc = 0.17). Once the equilibrium below is established, what is the equilibrium concentration of H* ions? HIO3 (aq) = H+ (aq) + 103- (aq) 7 8 9 x 100

Answer 1

Answer : The concentration of hydrogen ion at equilibrium is 0.118 M

Solution :  Given,

Concentration (c) = 0.200 M

Equilibrium constant = $k_c=0.17$

The equilibrium reaction is,

                          $HIO_3\rightleftharpoons H^++IO_3^-$

initially conc.       c           0         0

At eqm.          $c(1-\alpha)$     $c\alpha$        $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha)$.

Formula used :

$k_c=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha)$.

$0.17=\frac{(0.200\alpha)(0.200\alpha)}{0.200(1-\alpha)}$

By solving the terms, we get

$\alpha=0.59$

Now we have to calculate the concentration of hydrogen ion at equilibrium.

$[H^+]=c\alpha=0.200\times 0.59=0.118M$

Therefore, the concentration of hydrogen ion at equilibrium is 0.118 M