Question

Calculate the pH and the concentrations of all species present (H3O+, OH?, HIO3, and IO?3) in 0.0530M HIO3. Ka for HIO3is 1.7

Answer 1

HIO3 ionises as;

HIO3+H2O ------> IO₃^- + H3O+

At start, initial conc 0.0503   0   0

0.0503-x                  x         x

Ka= [ IO₃^- ] [H3O+]/ HIO3

1.7*10^-1 = x.x/ 0.0503-x

As conc of x is small, so can be neglected, 0.0503-x can be taken as 0.0503

1.7*10^-1 = x² / 0.0503

x² = 1.7*10^-1*0.0503

x= 9.24*10^-2

So, conc of IO3- = 9.24*10^-2 M

H+ = 9.24*10^-2 M

pH= -log[H+]= -log ( 9.24*10^-2)= 1.03

pH +pOH = 14

pOH= 14-1.03= 12.97

pOH= -log(OH-)

OH- = antilog (12.97) = 9.33*10^-12

conc of OH- = 9.33*10^-12M