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General Chemistry II An ionic salt is a compound comprised of a metal or ammoniu

ID: 568429 • Letter: G

Question

General Chemistry II An ionic salt is a compound comprised of a metal or ammonium ion and a nonmetal or polyatomic ion. When ionic salts dissolve in water they dissociate into ions. The more soluble the salt is the more ions will be present in its aqueous solution. To calculate the pH of a salt solution one needs to determine which one of the ions in solution can be a pH active species. 2. Consider a 0.45M aqueous solution of potassium chlorite. a) Write down the dissociation equation for this salt. Are there any spectator ions present in this solution? What are the pH active species in this solution? b) Write down the equilibrium reaction for this solution and then list all species present in the salt solution. c) Do you predict the pH of potassium chlorite solution to be above 7 or below 77 Above 7 EXPLAIN your choice: Below 7 d) Now fill in the ICE table below and calculate the pH for 0.45M solution of potassium Also please show that the 5% chlorite. Use exam appendix to find necessary approximation is valid in this case. Species Initial) Change [Equilibrium] Does the result match your prediction made in 2(c)?

Explanation / Answer

2. For 0.45 M KClO2

a) dissociation equation,

KClO2(aq) --> K+(aq) + ClO2-(aq)

K+ spectator ion

ClO2- is pH active species

b) ClO2- hydrolysis

      ClO2-(aq) + H2O(l) ---> HClO2(aq) + OH-(aq)

     base           acid          conjugate      conjugate

                                               acid              base

c) pH prediction for the solution

above 7

as OH- is generated which would increase the pH of the solution.

d) ICE chart

         ClO2-(aq) + H2O(l) ---> HClO2(aq) + OH-(aq)

I         0.45                -                  -                -

C          -x                 -                  +x             +x

E       0.45-x             -                    x               x

so,

Kb = [HClO2][OH-]/[ClO2-]

1 x 10^-14/1.1 x 10^-2 = x^2/(0.45-x)

x^2 + 9.1 x 10^-13x - 4.1 x 10^-13 = 0

x = [OH-] = 6.40 x 10^-7 M

pOH = -log[OH-] = 6.20

pH = 14 - pOH = 7.80

with 5% approximation,

Kb = [HClO2][OH-]/[ClO2-]

1 x 10^-14/1.1 x 10^-2 = x^2/(0.45)

x = [OH-] = 6.40 x 10^-7 M

pOH = -log[OH-] = 6.20

pH = 14 - pOH = 7.80

The result match with prediction made in 2(c)