Calculate the pH of each of the following solutions. (a) 0.22 M KNO2 (b) 0.25 M NaOCl (c) 0.58 M NH4ClO4?

Refer to: https://depts.washington.edu/eooptic/links/acidstrength.html
Ka(HNO₂) = 7.2 × 10⁻⁴
Ka(HOCl) = 2.9 × 10⁻⁸
Ka(NH₄⁺) = 5.8 × 10⁻¹⁰

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(a)
Kb(NO₂⁻) = Kw / Ka(HNO₂) = (1.0 × 10⁻¹⁴) / (7.2 × 10⁻⁴) = 1.39 × 10⁻¹¹

___________ NO₂⁻(aq) + H₂O(l) ⇌ HNO₂(aq) + OH⁻(aq) ____ Kb = 1.39 × 10⁻¹¹
Initial: ______ 0.22 M ___________ 0 M _____ 0 M
Change: ______ -y M ___________ +y M ____ +y M
Eqm: _____ (0.22 - y) M __________ y _______ y M

As Kb is very small, the dissociation of NO₂⁻ is to a very small extent.
Hence, we can assume that [NO₂⁻] at eqm = (0.22 - y) M ≈ 0.22 M

At eqm: Kb = [HNO₂] [OH⁻] / [NO₂⁻]
1.39 × 10⁻¹¹ = y² / 0.22
y = √{(1.39 × 10⁻¹¹) × 0.22} = 1.75 × 10⁻⁶

pOH = -log(1.75 × 10⁻⁶) = 5.8
pH = pKw - pOH = 14.0 - 5.8 = 8.2

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(b)
Kb(OCl⁻) = Kw / Ka(HOCl) = (1.0 × 10⁻¹⁴) / (2.9 × 10⁻⁸) = 3.45 × 10⁻⁷

___________ OCl⁻(aq) + H₂O(l) ⇌ HOCl(aq) + OH⁻(aq) ____ Kb = 3.45 × 10⁻⁷
Initial: ______ 0.25 M ___________ 0 M _____ 0 M
Change: ______ -y M ___________ +y M ____ +y M
Eqm: _____ (0.25 - y) M __________ y _______ y M

As Kb is very small, the dissociation of OCl⁻ is to a very small extent.
Hence, we can assume that [OCl⁻] at eqm = (0.25 - y) M ≈ 0.25 M

At eqm: Kb = [HOCl] [OH⁻] / [OCl⁻]
3.45 × 10⁻⁷ = y² / 0.25
y = √{(3.45 × 10⁻⁷) × 0.25} = 2.94 × 10⁻⁴

pOH = -log(2.94 × 10⁻⁴) = 3.5
pH = pKw - pOH = 14.0 - 3.5 = 10.5

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(c)
___________ NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) ____ Kb = 5.8 × 10⁻¹⁰
Initial: ______ 0.58 M ___________ 0 M _____ 0 M
Change: ______ -y M ___________ +y M ____ +y M
Eqm: _____ (0.58 - y) M __________ y _______ y M

As Ka is very small, the dissociation of NH₄⁺ is to a very small extent.
Hence, we can assume that [NH₄⁺] at eqm = (0.58 - y) M ≈ 0.58 M

At eqm: Ka = [NH₃] [H₃O⁺] / [NH₄⁺]
5.8 × 10⁻¹⁰ = y² / 0.58
y = √{(5.8 × 10⁻¹⁰) × 0.58} = 1.83 × 10⁻⁵

pH = -log(1.83 × 10⁻⁵) = 4.7