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mariarad [96]
1 year ago
14

Calculate the solubility of Ca(OH)2 in a 0.469M CaCl2 solution at 31°C, given that the Ksp of Ca(OH)2 is 4.96 x 10-6 at that tem

perature. Give your answer in milliMolar units 2.927 mM 1.626 mM 2.764 mM 1.382 mM 1.463 mM 3.252 mM​
Chemistry
1 answer:
Ghella [55]1 year ago
3 0

The solubility of Ca(OH)2 in a 0.469M CaCl2 solution at 31°C, at given Ksp of Ca(OH)2 is determined as 1.626 mM.

<h3>Dissociation reaction of Ca(OH)2</h3>

The dissociation reaction of Ca(OH)2 is given as follows;

Ca(OH)₂  ⇄ Ca²⁺  + 2OH⁻¹

                     x           2x

Concentration of Ca²⁺ = 0.469 M

Ksp = [x][2x]²

ksp = (0.469)(2x²)

ksp = 4(0.469)x²

ksp = 1.876x²

4.96 x 10⁻⁶ = 1.876x²

x² = (4.96 x 10⁻⁶)/(1.876)

x² = 2.643 x 10⁻⁶

x = √(2.643 x 10⁻⁶)

x = 1.626 x 10⁻³ M

x = 1.626 mM

Thus, the solubility of Ca(OH)2 in a 0.469M CaCl2 solution at 31°C, at given Ksp of Ca(OH)2 is determined as 1.626 mM.

Learn more about solubility here: brainly.com/question/23946616

#SPJ1

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Answer:

The answer is (H30+) =3,55e-8M and (OH-)=2,82e-7M

Explanation:

We use the formulas:

pH= - log(H30+)  and Kwater=(H30+)x(OH-)

pH= - log(H30+)  ----< (H30+)= antilog- pH=antilog- 7,45=3,55E-8M

Kwater=(H30+)x(OH-)

(OH-)=Kwater/(H30+)= 1,00e-14/3,55e-8 = 2,82e-7

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Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.

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Which list represents the classification of the elements nitrogen, neon, magnesium, and silicon, respectively?
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The question is incomplete. Complete question is:
Which list represents the classification of the elements nitrogen, neon, magnesium, and silicon, respectively?
(1) metal, metalloid, nonmetal, noble gas
(2) nonmetal, noble gas, metal, metalloid
(3) nonmetal, metalloid, noble gas, metal
(4) noble gas, metal, metalloid, nonmetal
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Answer: The correct answer is option 2) nonmetal, noble gas, metal, metalloid

Reason: 
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4) Metalloid have conductivity better than non-metals but poor than conductors. Hence, among the provided elements, Silicon satisfied the criteria. Hence, it is a metalloid.
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