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Andrei [34K]
3 years ago
10

A solution has a Pb2+ concentration of 1.9 x 10-3 M, and an F-1 concentration of 3.8 x 10-3 M. The value of Ksp for PbF2 at room

temperature is 4 x 10-8. Will this solution form a precipitate?
The solubility equilibrium equation is as follows.



.

Yes

No
Chemistry
2 answers:
horrorfan [7]3 years ago
6 0

A solution has a Pb2+ concentration of 1.9 x 10^-3, and an F-1 concentration of 3.8 x 10^-3 M. The value of Mps for Pbf2 at room temperature is 4 x 10^-8. This solution will NOT form a precipitate.
The solubility equilibrium equation is as follows.
PbF2 (s) <--> Pb2 (aq) + 2F- (aq)

valentina_108 [34]3 years ago
5 0
We are going to get the value of Qsp and compare it with the Ksp value:

when Qsp is the concentration of the initial products:

Qsp = [Pb2+][F-]^2

when the [Pb2+] = 1.9 x 10^-3

and [F-] = 3.8 x 10^-3

by substitution:

∴Qsp = (1.9 x 10^-3) * (3.8 x 10^-3)^2

           = 2.7 x 10^-8 

when Ksp = 4 x 10^-8

so Qsp < Ksp

∴ precipitate will not be formed 
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A cyanide solution with avolume of 12.73 mL was treated with 25.00 mL of Ni2+solution (containing
natima [27]

Answer:

The molarity of this solution is 0,09254M

Explanation:

The concentration of the Ni²⁺ solution is:

Ni²⁺ + EDTA⁴⁻ → Ni(EDTA)²⁻

0,03935L × 0,01307M = 5,143x10⁻⁴ moles Ni²⁺ ÷ 0,03010L =<em>0,01709M Ni²⁺</em>

25,00 mL of this solution contain:

0,01709M × 0,02500L = 4,2716x10⁻⁴ moles of Ni²⁺

The moles of Ni²⁺ that are in excess and react with EDTA⁴⁻ are:

0,01015L × 0,01307M = 1,3266x10⁻⁴ moles of Ni²⁺

Thus, moles of Ni²⁺ that react with CN⁻ are:

4,2716x10⁻⁴ - 1,3266x10⁻⁴ = 2,9450x10⁻⁴ moles of Ni²⁺

For the reaction:

4CN⁻ + Ni²⁺ → Ni(CN)₄²⁻

Four moles of CN⁻ react with 1 mole of Ni²⁺:

2,9450x10⁻⁴ moles of Ni²⁺ × \frac{4 mol CN^-}{1 molNi^{2+}} = <em>1,178x10⁻³ moles of CN⁻</em>

As the volume of cyanide solution is 12,73mL. The molarity of this solution is:

<em>1,178x10⁻³ moles of CN⁻ ÷ 0,01273L = </em><em>0,09254M</em>

I hope it helps!

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