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=((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=((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* 0.59=0.118M`

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