Chemistry: Explain how mixture melting points can be helpful in distinguishing between two compounds that have essentially the same melting point.

OH⁻ NH₃ HCO₃⁻ H₂O NH₄⁺ H₂CO₃ Explanation:We can do some rough calculations to find the approximate pH values of these solutions. H₂CO₃ Kₐ ≈ 10⁻⁶ NH₄⁺ Kb of NH₃ ≈ 10⁻⁵ Kₐ of NH₄⁺ ≈ 10⁻⁹ OH⁻ Strong base[OH⁻] = 10⁻¹ pOH = 1 pH = 14 - 1 = 13 HCO₃⁻ Salt of dibasic acid K₁ ≈ 10⁻⁶; K₂ ≈ 10⁻¹⁰ NH₃ Kb ≈ 10⁻⁵ pOH = 14 - 3 = 11 H₂O Neutral. pH = 7 Order from lowest [H₃O⁺] to highest [H₃O⁺]: OH⁻ NH₃ HCO₃⁻ H₂O NH₄⁺ H₂CO₃ pH 1 3 11 8 7 5 3.5 ...

Explain how mixture melting points can be helpful

Ответ:

naomicervero

14.07.2020

Chemistry

OH⁻ < NH₃ < HCO₃⁻ < H₂O < NH₄⁺ < H₂CO₃

Explanation:

We can do some rough calculations to find the approximate pH values of these solutions.

H₂CO₃

Kₐ ≈ 10⁻⁶

$\text{H}^{+} = \sqrt{K_{\text{a}}c} = \sqrt{10^{-6} \times 10^{-1}} = \sqrt{10^{-7}} = 10^{-3.5}\\\text{pH} = -\log (10^{-3.5}) = \mathbf{3.5}$

NH₄⁺

Kb of NH₃ ≈ 10⁻⁵

Kₐ of NH₄⁺ ≈ 10⁻⁹

$\text{H}^{+} = \sqrt{K_{\text{a}}c} = \sqrt{10^{-9} \times 10^{-1}} = \sqrt{10^{-10}} = 10^{-5}\\\text{pH} = -\log (10^{-5}) = \mathbf{5}$

OH⁻

Strong base

[OH⁻] = 10⁻¹

pOH = 1

pH = 14 - 1 = 13

HCO₃⁻

Salt of dibasic acid

K₁ ≈ 10⁻⁶; K₂ ≈ 10⁻¹⁰

${\text{H}^{+}} = \sqrt{K_{1}K_{2}} = \sqrt{10^{-6}\times 10^{-10}} = \sqrt{10^{-16}} = 10^{-8}\\\text{pH} = -\log (10^{-8}) = \mathbf{8}$

NH₃

Kb ≈ 10⁻⁵

$\text{OH}^{-} = \sqrt{K_{\text{b}}c} = \sqrt{10^{-5} \times 10^{-1}} = \sqrt{10^{-6}} = 10^{-3}\\\text{pOH} = -\log (10^{-3}) = 3$

pOH = 14 - 3 = 11

H₂O

Neutral. pH = 7

Order from lowest [H₃O⁺] to highest [H₃O⁺]:

OH⁻ < NH₃ < HCO₃⁻ < H₂O < NH₄⁺ < H₂CO₃

pH 1 3 11 8 7 5 3.5

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Explain how mixture melting points can be helpful in distinguishing between two compounds that have essentially the same melting point.

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