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Varvara68 [4.7K]
1 year ago
14

A mineral sample is analyzed for its cobalt and calcium content. A sample is dissolved, and then the cobalt and calcium are prec

ipitated as Co(OH)2(s) and Ca(OH)2(s) . At what pH can Co(OH)2(s) be separated from Ca(OH)2(s) at 25 °C? Assume that an effective separation requires a maximum concentration of the less soluble hydroxide of 1×10−6 M . Ksp[Co(OH)2]=5.9×10−15 M3 Ksp[Ca(OH)2]=5.0×10−6 M3
Chemistry
2 answers:
bekas [8.4K]1 year ago
7 0

Answer:

9.88

Explanation:

As higher is the Ksp, more soluble is the compound. So, Co(OH)₂ is the less soluble hydroxide.

The maximum concentration of it must be 1x10⁻⁶ M, and the reaction is:

Co(OH)₂(s) ⇄ Co⁺²(aq) + 2OH⁻(aq)

So, [Co⁺²] = 1x10⁻⁶M

Ksp =  [Co⁺²] *[OH⁻]²

[OH⁻]² = 5.9x10⁻¹⁵/1x10⁻⁶

[OH⁻] = √(5.9x10⁻⁹)

[OH⁻] = 7.6811x10⁻⁵

pOH = -log[OH⁻]

pOH = -log(7.6811x10⁻⁵)

pOH = 4.11

Knowing that pH + pOH = 14

pH = 14 - 4.11

pH = 9.88

Contact [7]1 year ago
5 0

Co(OH)₂ can be separeted from Ca(OH)₂ at <u>pH : 9.885</u>

<h3>Further explanation </h3>

Solubility is the maximum amount of a substance that can dissolve in some solvents.

Ksp is an ion product in equilibrium

Solubility (s) and solubility constants (Ksp) of the AxBa solution can be stated as follows.

AₓBₐ (s) ← ⎯⎯⎯⎯ → aAᵃ⁺ (aq) + b Bᵇ⁻ (aq)

s                             as               bs

Ksp = [Aᵃ⁺]ᵃ [Bᵇ⁻]ᵇ

Ksp = (as)ᵃ (bs)ᵇ

Solubility units in the form of mol / liter or gram / liter

Ksp is the product of ions in an equilibrium saturated state

Qsp is the product of the ion ions from the reacting substance

Qsp <Ksp = solution has not occurred sediment

Qsp = Ksp = saturated solution, exactly the precipitate will occur

Qsp> Ksp = sediment solution

The smaller the solubility of a substance, the more difficult it is to dissolve, the smaller the Ksp value. The greater the Ksp, the greater the solubility so that it can dissolve more easily

From the 2 hydroxide solutions above, Co(OH)₂ has a smaller Ksp value, so that a precipitate will form first so that it can be separated from the Ca(OH)₂ solution

Because the Co concentration is 1 × 10−6 M, we determine the pH when  Co(OH)₂ precipitated

Co(OH)₂ --> Co²⁺ +2OH⁻

Ksp Co (OH)₂= [Co²⁺][OH⁻]²

5.9.10⁻¹⁵  = 1.10⁻⁶[OH⁻]²

\displaystyle [OH^-]^2=\frac{5.9.10^{-15}}{1.10^{-6}}

[OH^-]=7.68.10^{-5}

\displaystyle pOH=-log~7.68.10^{-5}\\\\pOH=4.114\\\\pH=14-4.114=\boxed{\bold{9.885}}

<h3>Learn more</h3>

the rate of solubility

brainly.com/question/9551583

influencing factors are both solubility and the rate of dissolution

brainly.com/question/2393178

The solubility of a substance

brainly.com/question/2847814

Keywords: solubility, Ksp, precipitated, less soluble hydroxide, Co(OH)₂,Ca(OH)₂

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Consider the following hypothetical reaction: 2 P + Q → 2 R + S The following mechanism is proposed for this reaction: P + P Q
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3 0
1 year ago
0.9775 grams of an unknown compound is dissolved in 50.0 ml of water. Initially the water temperature is 22.3 degrees Celsius. A
elena-14-01-66 [18.8K]

Answer:

The enthlapy of solution is -55.23 kJ/mol.

Explanation:

Mass of water = m

Density of water = 1 g/mL

Volume of water = 50.0 mL

m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g

Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C

Heat capacity of water,c =4.186 J/g°C

Heat gained by the water when an unknown compound is dissolved be Q

Q= mcΔT

Q=50.0 g\times 4.186 J/g^oC\times 4.7^oC=983.71 J

heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.

Q'=-Q

Q'= -983.71 J =-0.98371 kJ

Moles of unknown compound = \frac{0.9975 g}{56 g/mol}=0.01781 mol

The enthlapy of solution :

\frac{Q'}{moles}

=\frac{-0.98371 kJ}{0.01781 mol}=-55.23 kJ/mol

The enthlapy of solution is -55.23 kJ/mol.

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