The molar mass of gas = 238.29 g/mol
<h3>Further explanation</h3>Given
mass = 81.5 g
P=1.75 atm
V=4.92 L
T=307 K
Required
molar mass
Solution
The gas equation can be written
So the equation becomes :
Input the value :
What are valence electrons
1 answer:
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Answer:
The answer to the question is;
The first ion to precipitate out is the Al³⁺ ion and the concentration of the Al³⁺ ion when the Ca²⁺ ion begins to precipitate is 1.12 × 10⁻⁵ M.
Explanation:
To solve the question, we note that
aluminum nitrate, Al(NO₃)₃ will dissociate as follows
Al(NO₃)₃ → Al³⁺ (aq) + 3NO₃⁻ (aq)
Therefore when sodium phosphate is added to a solution that contains aluminum nitrate we have the following system of aluminium phosphate which is
AlPO₄(s) ⇄ Al³⁺(aq) + PO₄³⁻(aq)
The solubility product for the above reaction is
Ksp = [Al³⁺][PO₄³⁻] = 9.84×10⁻²¹
The solubility product for calcium phosphate is expressed as
Ca₃(PO₄)₂(s) ⇄ 3 Ca²⁺(aq) + 2 PO₄³⁻(aq)
With Ksp = [Ca²⁺]³[PO₄³⁻]² = 2.07×10⁻³³
From the solubility product, we can find the concentration of [PO₄³⁻] at which precipitation starts as follows
The phosphate concretion for Al³⁺ when precipitation starts is
[PO₄³⁻] = = 9.84×10⁻²¹ / 0.007 = 1.406×10⁻¹⁸ M
The phosphate concretion for Ca²⁺ when precipitation starts is
[PO₄³⁻] = =
= 8.75×10⁻¹⁶ M
(Aluminium phosphate precipitates out first)
The reaction favors the precipitation of the aluminum phosphate first due to the lower concentration of the [PO₄³⁻] ions in the [Al³⁺][PO₄³⁻] system which is lower than the relative [PO₄³⁻] in the [Ca²⁺]³[PO₄³⁻]².
Therefore, the more sodium phosphate added serves to precipitate the remaining aluminium phosphate.
The process continues and the concentration of Al³⁺ decreases as more precipitates form. The process continues until the equilibrium conditions satisfies the precipitation threshold level for the calcium phosphate system concentration whereby the concentration of the Al³⁺ in the solution is given by.
[Al³⁺] = = 1.12 × 10⁻⁵ M
Therefore the concentration of this aluminium ion when the calcium ion begins to precipitate = 1.12 × 10⁻⁵ M.