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

Answer:

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