1) Write the hydrolysis reaction(s) for the salt, KClO, and determine if the solution is acidic, basic, or neutral.

Question

2) A solution contains 0.020 M KClO, what is the pH of the solution? Ka (HClO) = 3.5x10-8

Answer 1

The hydrolysis reaction can be written ClO^1-  + H2O → HClO + OH^1-

K_H  = [HClO] [OH^1-] / [ClO^1-] where K_H = K_w / K_a = 10^-14 / 3.5 x 10^-8 = 2.86 x 10^-7.

Since [HClO] =  [OH^1-], we can write Kh = [OH-]^2 / {[ClO^1-] – [OH-]} (equation 1)

Since [OH-] is << 0.02, we can neglect the [OH-] in the denominator of equation 1.

Now we have Kh = [OH-]^2 / 0.02.

Solve for [OH-] to get 7.56 x 10^-5. [H+] = 10^-14 / 7.56 x 10^-5 = 1.32 x 10^-10.

pH = - log 1.32 x 10^-10 = 9.9

Check the math.

Answer 2

(1) KClO is a strong electrolyte. It will completely dissociate into K⁺ and ClO^- when dissolving water. K⁺ is from strong base, KOH and will not affect the pH of the solution where water is the solvent. On the other hand, ClO^- is from weak acid (HClO) and will affect the pH of the solution. The hydrolysis of ClO^- will produce additional OH^- to make the solution weakly basic. The equations are as follows.

KClO → K⁺ + ClO^- (dissociation of KClO in water, reaction goes to completion, so single arrow is used.)

ClO^- + H₂O ↔ HClO + OH^- (hydrolysis of ClO^-, reaction cannot go to completion. Instead it reaches equilibrium at certain point. So double arrow is used)

(2) pH of 0.02 M KClO is calculated as follows

Step 1: Determine the concentration of OH^- based on ICE table.

ClO^- + H₂O ↔ HClO + OH^-

                                         Initial  0.020 M              0 M    0 M

                                           Change  – X M              + X M + X M

                              Equilibrium   (0.020 – X) M              X M    X M

At equilibrium, ionization constant of the weak base ClO^-, K_b is written as

Kb = [HClO] [OH^-]/[ClO^-] = X × X /(0.020 – X) = X²/(0.020– X) = Kw/Ka = 1.0 x 10^-14/3.5 × 10^-8 = 2.9 × 10^-7. Here an important relation between Ka (ionization constant of acid), Kb (ionization constant of the conjugate base) and Kw (1.0 x 10^-14, the ion product constant of water at 25 ^oC) is applied. Ka × Kb = Kw

Then one assumption can be used to make the life easier. As Kb is very small, it is true that X is much less than the initial concentration 0.020 M. So 0.020 – X ≈ 0.020.

So we have the simplified equation as follows.

X²/0.020 = 2.9 × 10^-7.

X² = 2.9 × 10^-7 × 0.020 = 5.8 × 10^-9

X = 7.6 × 10^-5 M

Step 2

The above X is also the concentration of OH^-, or [OH^-].

pOH = –log[OH^-] = –log7.6 × 10^-5 = 4.1

So pH = 14.0 – pH = 14.0 – 4.1 = 9.9

Here the relation between pH and pOH is applied. pH + pOH = 14.0

the pH (9.9) confirms the conclusion from the Question 1: the solution is weekly basic.

Hope this helps. Let me know if you have any questions.

Answer 3

KClO can be made from:

KOH + HClO -> KClO + H₂O

KOH is a strong base while HClO, as we are given, is a weak acid with Ka = 3.5x10^-8.

We know that the pH of a salt dissolved in water made from a strong base & weak acid should be basic, so our result should have pH > 7.

The starting hydrolysis reaction is:

KClO + H₂O -> K⁺ + HClO^- + OH^-

K⁺ is a spectator ion, it will not impact the acid-base equilibrium.

So we rewrite the hydrolysis to include only relevant species as:

ClO^- + H₂O <-> HClO + OH^-

Note we are producing hydroxide ions, as we mentioned earlier, the pH should be above 7 in the end.

Make an ICE table to calculate the ending concentrations of the species:

ClO^- + H₂O <-> HClO + OH^-

0.02 0 0

-x +x +x

---------------------------------------

0.02-x x x

x = [OH^-] = [HClO]

We are given the Ka of HClO, however the expression we have above is that of the Kb

general form:

Kb = [conjugate_acid][OH^-]/[base]

With our species:

Kb = [HClO][OH^-]/[ClO^-]

We know Kw = Ka*Kb

Kw is the water dissociation constant; Kw = [H+][OH-] = 1x10^-14 (at 25^oC)

Kb = Kw/Ka = 1x10^-14 / 3.5x10^-8 = 2.9x10^-7

Use Kb and ICE table to calculate x, the [OH^-].

Kb = x*x / (0.02-x) = x² / (0.02-x)

Assume x is very small compared to the initial 0.02M starting concentration, as is typical for weakly dissociating acids/bases

Kb = x² / 0.02

x² = 0.02 * Kb

x = sqrt(0.02 * Kb) = sqrt(0.02 * 2.9x10^-7)

x = [OH^-] = sqrt(0.02 * 2.9x10^-7)

[OH^-] = 7.6x10^-5 M

We can use [OH^-] to find [H⁺] from which we can calculate pH = -log([H⁺])

Kw = 1x10^-14 = [H⁺][OH^-]

[H⁺] = (1x10^-14)/[OH^-]

[H⁺] = (1x10^-14)/(7.6x10^-5) = 1.3x10^-10 M

pH = -log(1.3x10^-10 M)

pH = 9.9

This pH is basic, above 7, as should be the case for salts derived from weak acids & strong bases