Question

Order these species by increasing concentration of H30+ in a 1.0 M aqueous solution. (From the

solution with the least hydronium concentration to the solution with the most hydronium concentration)
NO
H2CO3, NH4, OH, HCO3, NH3, H20
Home
ir
H2CO3,NH4+, OH", HCO3, NH3,
H20
Paste
H20, H2CO3, NH4+, OH", HCO3-
NH3
6
con
O
OH", NH3, HCO3, H20, NH4+,
H2CO3
None of the answer choices are
correct.

Answer 1

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

Step-by-step 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) * 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) * 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)* 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) * 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