Solubility product constants are values to describe the saturation of ionic compounds with low solubility. A saturated solution is when there is a dynamic equilibrium between the solute dissolved, the dissociated ions, the undissolved and the compound. It is calculated from the product of the ion concentration in the solution. For barium chromate, the dissociation would be as follows:
BaCrO4 = Ba^2+ + (CrO4)^2-
So, the expression for the solubility product would be:
Ksp = [Ba^2+] [(CrO4)^2-]
we let x = [BaCrO4] = [Ba2+] = [(CrO4)2-] = 2.81x10^-3 g/L ( 1 mol / 253.35 g ) = 1.11x10^-5
Ksp = x(x)
Ksp= x^2
Ksp = (1.11x10^-5)^2
Ksp = 1.23x10^-10
The Ksp of Barium chromate at that same temperature for the solubility would be 1.23x10^-10.
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Explanation:
The value of current solubility product is calculated as below
K = (Ag+)( Cl-)
Ag+ = 0.01 M
Cl-= 1 x10^-5M
K is therefore = 1 x10^-5 x 0.01 = 1 x10 ^ -7 M
The K obtained is greater than Ksp
that is K> KSp
1x10^-7 > 1.7 x10 ^-10
will precipitation of AgCl form?
yes the precipitation of AgCl will be formed since K> KSP
Answer:
100.432ºC
Explanation:
Ebullioscopy is the elevation of the boiling point of a solvent that has a solute nonvolatile dissolved. The change in the temperature may be calculated by Raoult's law:
ΔT = Kb.W.i
Where <em>ΔT</em> is the temperature change, <em>Kb</em> is the ebullioscopy constant,<em> W</em> is the molality and <em>i</em> is the Van't Hoff factor, which determinates the particles that affect the proper.
The molality is:
W = m1/(M1xm2)
Where <em>m1</em> is the solute mass (in g), <em>M1</em> is the molar mass of the solute, and <em>m2</em> is the mass of the solvent (in kg).
The Van't Hoff factor is the number of final particles divided by the number of initial particles. Magnesium is from froup 2, so it forms the cation Mg⁺², and chlorine ins from group 7 and forms the anion Cl⁻, the salt is MgCl₂ and dissociates:
MgCl₂ → Mg⁺²(aq) + 2Cl⁻(aq)
So, it has 1 particle in initial, and 3 in final (1 Mg⁺² and 2 Cl⁻). So:
i = 3/1 = 3.
The molar mass of MgCl₂ is: 24.3 + 2x35.5 = 95.3 g/mol, m1 = 6.69g, m2 = 0.250 kg, so:
W = 6.69/(95.3x0.250)
W = 0.281 m
Then:
ΔT = 0.512x0.281x3
ΔT = 0.432ºC
The normal boling point of water is 100ºC, so
T - 100 = 0.432
T = 100.432ºC
Answer:
Compare the solubility of silver iodide in each of the following aqueous solutions:
a. 0.10 M AgCH3COO
b. 0.10 M NaI
c. 0.10 M KCH3COO
d. 0.10 M NH4NO3
1. More soluble than in pure water.
2. Similar solubility as in pure water.
3. Less soluble than in pure water.
Explanation:
This can be explained based on common ion effect.
According to common ion effect the solubility of a sparingly soluble salt decreases further in a solution which has a common ion to it.
The solubility of AgI(s) silver iodide in water is shown below:
a. a. 0.10 M AgCH3COO has a common ion Ag+ with AgI.
So, AgI is less soluble than in pure water in this solution.
b. 0.10 M NaI has a common ion I- with AgI.
So, AgI is less soluble than in pure water in this solution.
c. 0.10 M KCH3COO:
This solution has no common ion with AgI.
So, AgI has similar solubility as in pure water.
d. 0.10 M NH4NO3:
In this solution, AgI can be more soluble than in pure water.
