Calculate the pH of each of the following solutions. (a) 0.12 M KNO2 WebAssign will check your answer for the correct number of significant figures. (b) 0.37 M NaOCl WebAssign will check your answer for the correct number of significant figures. (c) 0.39 M NH4ClO4 WebAssign will check your answer for the correct number of significant figures.

a. pH → 8.12

b. pH → 10.55

c. pH → 8.38

Explanation:

In this excersise, the first two salts come from a strong base and a weak acid, so the anion will make hydrolisis that can help us, to determine pH.

The  NH₄ClO₄ comes from a weak acid and a weak base. This is an special case.

a. KNO₂ →  K⁻  +  NO₂⁻

NO₂⁻  +  H₂O  ⇄  HNO₃  +  OH⁻        Kb = 1.41×10⁻¹¹

Kb = x² / (0.12 - x)

We can avoid the x in denominator because Kb it is a very low value. This expression is reduced to:

[OH⁻] = √(Kb . 0.12)

[OH⁻] = √(1.41×10⁻¹¹ . 0.12)  = 1.30×10⁻⁶ M

- log [OH⁻] = pOH → - log 1.30×10⁻⁶  = 5.88

14 - pOH = pH → 14 - 5.88 = 8.12

b. NaOCl  →  Na⁻  +  OCl⁻

OCl⁻  +  H₂O  ⇄  HClO  +  OH⁻        Kb = 3.33×10⁻⁷

The same as before, we avoid the x in denominator because Kb it is a very low value.

[OH⁻] = √(Kb . 0.37)

[OH⁻] = √(3.33×10⁻⁷ . 0.37)

[OH⁻] = 3.51×10⁻⁴ M

- log 3.51×10⁻⁴ = 3.45 → pOH

14 - 3.45 = 10.55 → pH

c. NH₄ClO₄ → NH₄⁺  +  ClO⁴⁻

Both ions can make hydrolisis in water. Ammonium can give ammonia and the hypochlorite makes hypochlorous acid.

NH₄⁺  +  H₂O  ⇄  NH₃  +  H₃O⁺      Ka = 5.7×10⁻¹⁰

OCl⁻  +  H₂O  ⇄  HClO  +  OH⁻        Kb = 3.33×10⁻⁷

The reduced expression to calculate pH for this sort of salts, is to make the square root of the product between the Ka constant.

As we have Kb for the HClO, we need to determine the Ka

Kw = Ka . Kb   →  Kw/Kb = Ka → 1×10⁻¹⁴ / 3.33×10⁻⁷ = 3×10⁻⁸

[H₃O⁺] = √(5.7×10⁻¹⁰ . 3×10⁻⁸) = 4.14×10⁻⁹ M

- log [H₃O⁺] = pH → - log 4.14×10⁻⁹ = 8.38

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