Question

1) Calculate the pH and the concentrations of all species present (H3O+, OH−, HIO3, and IO−3) in 0.0528 M HIO3. Ka for HIO3 is 1.7×10−1.

2) Calculate the concentrations of all species present (H3O+, OH−, HIO3, and IO−3) in 0.0528 M HIO3.

Answer 1

Solution :-

Calculate the pH and the concentrations of all species present (H3O+, OH−, HIO3, and IO−3) in 0.0528 M HIO3. Ka for HIO3 is 1.7×10−1

HIO3    +      H2O ------ >   H3O+      +       IO3-

0.0528 M                            0                    0

-x                                         +x                 +x

0.0528-x                               x                   x

Ka= [H3O+][IO3-]/[HIO3]

1.7*10^-1 = [x][x]/[0.0528-x]

1.7*10^-1 * 0.0528-x = x^2

Solving for the x we get

0.04228 M= x

So the concnetrations of the each species are as follows

[HIO3] = 0.0528 M – 0.04228 M = 0.0105 M

[H3O+] = x= 0.04228 M

[IO3-] = x = 0.04228 M

HO- = 1*10^-14 / 0.04228 = 2.36*10^-13 M

pH of the solution is calculated as

pH= -log [H3O+]

pH= -log [0.04228]

pH= 1.37