Respuesta :
Answer:
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
As per the calculations of the pH the solutions of 0.12 M KNO2 and 0.37 M NaOCl and the 0.39 M NH4ClO4 web assign have been made with a significant check on the figures.
- Thus the answer for the first is pH → 8.12 and that of second calculation is pH → 10.55 and the last id pH → 8.38.
Learn more about the Calculate the pH of each of the following solutions.
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