Now for a more challenging problem, let\'s look at an ionic salt; An ionic salt

Question

Now for a more challenging problem, let's look at an ionic salt; An ionic salt is generally thought of as a compound comprised of a metal and a nonmetal or a polyatomic ion. Ionic salts dissociate into ions in aqueous solution. One of these ions is almost always a pH active species and it is this species that we would use to do pH calculations. Consider an aqueous solution of potassium chlorite (KClO_2). Identify the spectator ion and the pH active species in this solution? Write the equilibrium reaction for this solution. Remember that you consider ONLY the pH active species when you do this. Do you predict the pH of KClO_2 to be above 7 or below 7? Above 7 Below 7 EXPLAIN your choice: Given that the K_a of HClO_2 = 1.1 times 10^-2, calculate the pH for a 0.45 M solution of KClO_2. You might need to review the equations provided in part 2 to solve this problem.

Explanation / Answer

Q3.

(A) KClO2 dissociats as:

KClO2 -------> K+ + ClO2-

The spectator ion is K+ and ClO2- is pH active ion.

(B) The equilibrium equation of pH active species is given below.

ClO2- + H2O <------->HClO2 + OH-

(C) The pH will be above 7. This is because the ClO2- produces OH- ions, which make solution basic.

(d) ClO2- + H2O <------->HClO2 + OH-

I 0.45 0 0

C -x +x +x

E 0.45 - x x x

Ka X Kb = Kw ; Kb = Kw/Ka = 10-14 / 1.1 X 10-2 = 0.9 X 10-12 = 9.0 x 10-13

9.0 x 10-13 = x2 / (0.45 - x)

Above equation will attain the form of following quadratic equation

x2 + 9.0X10-13x - 9.0X10-13 = 0

x = 9.49 X 10-7

or OH- = 9.49 X 10-7

[H+] [OH-] = 10-14

[H+] = 10-14 / 9.49 X 10-7 = 1.05 x 10-8

pH = -log[H+] = -log[1.05 x 10-8] = 7.97

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