How does high pressure effect the solubility of a gas?
2 answers:
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3.08 is the pOH of a 0.175 M aqueous solution of .
0.215% is the per cent ionization of a 0.325 M aqueous solution of
pH is a measure of how acidic/basic water is.
A)
→
Kb = 4.5 x
Kb = {concentration of (NH₄⁺) x concentration of (OH⁻)} ÷ concentration of (NH₃).
concentration of (NH₄⁺) = concentration of (OH⁻) = x.
x² = Kb x concentration of (NH₃)
x² = 4.5 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.
x = concentration of (OH⁻) = √(7.0 × 10⁻⁷)
= 8.367 × 10⁻⁴
pOH = -log(c(OH⁻))
=- log ( 8.367 × 10⁻⁴)
= 3.08
B)
Chemical reaction: NX₃ + H₂O ⇄ NX₃H⁺ + OH⁻.
Concentration of (NX₃) = 0.325 M.
Kb = 4.5 x 10⁻⁶.
[NX₃H⁺] = [OH⁻] = x.
[NX₃] = 0.325 M - x.
Kb = [NX₃H⁺] x [OH⁻] ÷ [NX₃].
4.5 x 10⁻⁶ = x² ÷ (0.325 M - x).
x = 0.0007 M.
Per cent of ionization:
α = 0. 0007 M ÷ 0. 325 M x 100%
= 0.215%.
Hence,
3.08 is the pOH of a 0.175 M aqueous solution of .
0.215% is the per cent ionization of a 0.325 M aqueous solution of
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Mass of Fe₂O₃ : 74.26 g
Volume of O₂ at STP = 15.624 L
<h3>Further explanation</h3>Reaction
4Fe (s) + 3O₂ (g) ==> 2Fe₂O₃ (s)
mol Fe :
Ar Fe = 56 g/mol
mol Fe₂O₃ :
mass Fe₂O₃ :
MW Fe₂O₃ : 159,69 g/mol
mol O₂ :
volume O₂ at STP(1 mol=22.4 L)
Given :
0.00072 M solution of at
.
To Find :
The concentration of and pOH .
Solution :
1 mole of gives 2 moles of
ions .
So , 0.00072 M mole of gives :
Now , pOH is given by :
Hence , this is the required solution .