What is the molar solubility (M) of PbBr2 in pure water? What is the molar solubility (M) of PbBr2 in 0.500 M KBr
the solution? What is the molar solubility (M) of PbBr2 in 0.500 M Pb(NO3)2
the solution?
Answer 1
general orientation
Concepts and reason
Solubility:
Solubility is defined as the maximum amount of solute dissolved in a given amount of solvent to obtain a saturated solution at a given temperature.
Molar solubility:
Molar solubility is the number of moles of a solute that can be dissolved in one liter of solution. It is expressed in mol/L or M (molarity).
Common ionic effect:
The common ion effect is defined as the solubility of a partially soluble salt that will decrease with the addition of a soluble salt that has an ion in common with it.
Fundamentals
Consider a general reaction:
The relationship between the solubility product and the molar solubility is as follows:
Where,
Solubility product =
Molar solubility of ion M =
Molar solubility of ion X =
Step by step
Step 1 of 3
Therefore, the molar solubility (M) of in pure water is.
The given compound is slightly soluble in water and the solubility product constant is . The molar solubility (M) of in pure water is calculated using the expression of the solubility product.
For example, it is an error. Because one mole of solid dissociates into one mole of and two moles of . The correct equation is therefore .
Find the molar solubility (M) of a 0.500 M solution.
Step 2 of 3
Therefore, the molar solubility (M) of in is.
The molar solubility (M) of in is calculated using the expression of the solubility product replacing the common ion concentration. The solubility of the slightly soluble salt is decreased by addition to the solution. Therefore, the molar solubility (M) of in is lower than that of water.
Not to be confused with molar solubility in pure water and in KBr solution. The molar solubility value is decreased by adding KBr due to the common ion effect.
Find the molar solubility (M) of a 0.500 M solution.
Step 3 of 3
Therefore, the molar solubility (M) of in is .
The molar solubility (M) of in pure water is.
The molar solubility (M) of in is.
The molar solubility (M) of in is.
The molar solubility (M) of in is calculated using the expression of the solubility product replacing the common ion concentration. The solubility of the slightly soluble salt is decreased by addition to the solution. Therefore, the molar solubility (M) of in is lower than that of water.
Answer
The molar solubility (M) of in pure water is.
The molar solubility (M) of in is.
The molar solubility (M) of in is.
answer only
The molar solubility (M) of in pure water is.
The molar solubility (M) of in is.
The molar solubility (M) of in is.
MX) NM”* (aq) +mX”+ (aq)
K = [mm*]” [x”]
M
Let: K, = 6.60×106 PbBr, (s) Pb2+ (aq) + 2Br” (aq) K. =[Pb2+ ][Br] Assume:[Pb2+] =if[Br]=s Substitution, 6.60×10* =sx(25) 45° = 6.60×10 Solve for, s=1.18x10PM
1.18×10²M
.01×099
[Pb2+ ][Br]
[Pb2+ ] =2[Br]
Data: K, = 6.60×106 [Br] =0.500 M PbBr2(8) Pb?* (aq) +2Br” (aq) KR = [Pb2+ ][Br] Assume: [Pb2+ ]=s
Substitution, 6.6010% =sx(0.500) 6.60×106 (0.500) Solve for, s=2.64x10PM
2.64x10SM
(ON)
Given, Kop = 6.60~10% Pb2+]=0.500M
PbBry() Pb2+ (aq) + 2Br (aq) K.=[Pb²+ ][Br] Presume, [Br”]=s
Substitution, 6.60~10= 0.500x(25) 62 – 6.60*10 4×0.500 s=1.82×10-‘M
(ON)
1.82×10′M
(ON)
[РЫ?]
(ON)
(ON)
1.18×10²M
2.64x10SM
(ON)
1.82 × 10′ M
1:18 * 10:00 p.m.
2.64x10SM
(ON)
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