Question

QUESTION 3 Calculate the pH of a solution that is 0.419 M HCIO and 0.422 M KCIO. Ka of HCIO is 3.0 x 10-8

Answer 1

Given, concentration of HClO = 0.419M

    concentration of KClO = 0.422M

Ka of HClO = 3.0 x 10-8

pH of solution containing 0.419M HClO and 0.422M KClO = ?

by using ICE table method, we can calculate the pH of solution containing 0.419M HClO and 0.422M KClO.

HClO --------------------> H+ + ClO-

Initial conc. 0.419M 0 0

Change con.c   0.419-x x 0.422 + x ( Here KClO dissolves almost completely & gives 0.422M ClO-)

End conc. 0.419 x    0.422 (here, x is very small comparatively with 0.419 & 0.422 therefore, we ignored the "x".

But we know that for the above equation, Ka = (0.422)(x)/(0.419) ----- eqn(1)

but, given Ka= 3.0 x 10-8

therefore eqn.(1) becomes, 3.0 x 10-8 = (0.422)(x)/(0.419)

therefore, x=3.0 x 10-8 x (0.419)/(0.422)

x= 3.0213 x 10-8

but, x = [ H+ ] and thus, [ H+ ] = 3.0213 x 10-8

from above calculations, we get the concentrations of H+ in a given solution is  3.0213 x 10-8  

from definition of pH= -log[ H+ ] = -log[3.0213 x 10-8] = 7.52

Answer: pH of solution containing 0.419M HClO and 0.422M KClO is 7.52