Question

Calculate the concentrations of all five chemical species present in a 0.750 M solution of hypoiodous acid.

Answer 1

Solution :-

Reaction equation

HiO(aq) + H2O(l) ------ > H3O^+(aq) + IO^-(aq)

ka of the HIO = 3.2*10^-11

so lets calculate the dissociation concnetration of the ions

          HiO(aq) + H2O(l) ------ > H3O^+(aq) + IO^-(aq)

I        0.750 M                                0                0

C        -x                                         +x             +x

E      0.750 -x                                 x                x

ka = [H3O^+] [ IO^-] / [HI]

lets put the values in the formula

3.2*10^-11= [x][x]/[0.750-x]

since ka is very small therefore neglect the x from the denominator then we get

3.2*10^-11= [x][x]/[0.750]

3.2*10^-11 * 0.750 = x^2

2.4*10^-11 = x^2

taking square rrot of both sides we get

4.9*10^-6 M= x

therefore the concentration of the [H3O^+] = 4.9*10^-6 m

[IO^-] = 4.9*10^-6 M

[OH-] = kw/[H3O+]

[OH-] = 1*10^-14 / 4.9*10^-6 =2.04*10^-9 M

[HIO] = 0.750 M - 4.9*10^-6 M = = 0.750 M