Calculate the molar solubility of Cd(OH)2 when buffered at a pH = 12.30. The Ksp for Cd(OH)2 is 2.5 x 10-14. Calculate the molar solubility of Cd(OH)2 when buffered at a pH = 12.30. The Ksp for Cd(OH)2 is a.2.5 x 10-14 M.
b. 8.5 x 10-6 M
c. 6.3 x 10-11 M
d. 1.3 x 10-12 M
e. 5.0 x 10-2 M
f. 1.8 x 10-5 M

Answers

Answer 1

Answer:

Answer:

c. 6,3x10⁻¹¹M

Explanation:

The solubility of a buffer is defined as the concentration of the dissolved solid in a saturated solution. For the Cd(OH)₂, solubility is:

[Cd²⁺] = S

The dissolution of Cd(OH)₂ is:

Cd(OH)₂ ⇄ Cd²⁺ + 2OH⁻

And the ksp is defined as:

ksp = [Cd²⁺][OH⁻]²

As ksp = 2,5x10⁻¹⁴ and [OH⁻] at pH=12,30 = 10^-(14-12,30) = 0,01995M

2,5x10⁻¹⁴ = [Cd²⁺]×(0,01995M)²

[Cd²⁺] = 6,3x10⁻¹¹M

That means solubility is c. 6,3x10⁻¹¹M

I hope it helps!

Answer 2

Answer:

Final answer:

The molar solubility of Cd(OH)2 when buffered at a pH of 12.30 can be calculated using the concept of hydrolysis. The correct answer is 6.3 x 10^(-11) M.

Explanation:

To calculate the molar solubility of Cd(OH)2 when buffered at a pH of 12.30, we need to use the concept of hydrolysis. Cd(OH)2 is a slightly soluble salt that undergoes hydrolysis in aqueous solution. At a high pH value, OH- ions react with water to form more OH- ions, shifting the equilibrium towards the hydrolysis reaction.

First, we write the balanced equation for the hydrolysis reaction: Cd(OH)2(s) ⇌ Cd2+(aq) + 2OH-(aq)
Since OH- is being produced, we can assume that the concentration of OH- is much greater than that of Cd2+. Therefore, we can ignore the concentration of Cd2+ when calculating the solubility product (Ksp).
Next, we use the equation for the hydrolysis reaction to write the expression for the solubility product constant (Ksp): Ksp = [Cd2+][OH-]^2
The concentration of OH- ions in a basic solution is related to the pH by the equation: pOH = 14 - pH
Using this equation, we can calculate the pOH of the buffered solution: pOH = 14 - 12.30 = 1.70
Then, we convert the pOH back to OH- concentration: [OH-] = 10^(-pOH) = 10^(-1.70)
Finally, we substitute the calculated [OH-] into the expression for Ksp to solve for the molar solubility of Cd(OH)2: [Cd(OH)2] = sqrt(Ksp / [OH-]^2)

After performing the calculations, the molar solubility of Cd(OH)2 when buffered at a pH of 12.30 is approximately 6.3 x 10^(-11) M. Therefore, the correct answer is option c. 6.3 x 10^(-11) M.

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Further Explanation

Chemical equation

A chemical equation is an equation showing chemical symbols of reactants and those of products. They represent a chemical reactions between reactants to form products.
For example; (NH4)2S(aq)+SrCl2(aq)→ 2 NH4Cl(aq) + SrSO4(s), where (NH4)2S and SrCl2 are reactants while NH4Cl and SrSO4 are products.

Types of chemical reactions

Precipitation reaction

Precipitation reactions are reactions which involves the formation of a precipitate as one of the products. A precipitate is a compound that is insoluble in water.
An example of a precipitation reaction is; (NH4)2S(aq)+SrCl2(aq)→ 2 NH4Cl(aq) + SrSO4(s), where the compound SrSO4 is the precipitate.

Displacement reaction

Displacement reactions are reactions where ions replace other ions in their compounds.
For example; ; (NH4)2S(aq)+SrCl2(aq)→ 2 NH4Cl(aq) + SrSO4(s) is an example of a double displacement reaction where NH4+ takes the place of Sr ions in SrCl2 and Sr2+ takes the place of NH4+ in (NH4)2SO4.

Decomposition reaction

Decomposition reactions are reactions which involves break down of a compound to its constituent’s elements or other compounds by use of a catalyst or heat.
For example; Decomposition of lead (II) nitrate using heat to get lead (ii) oxide, oxygen and nitrogen (IV) oxide.

Neutralization reaction

Neutralization reactions are reactions that involve reacting a base or an alkali and an acid to form a salt and water as the only product.

Redox reactions

Redox reactions are reactions that involve both reduction and oxidation. Some species in reactions undergo reduction while others undergo oxidation.

Keywords: Chemical reactions, precipitation reactions, chemical equations

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Level: High school

Subject: Chemistry

Topic: Chemical reactions

Sub-topic: Precipitation reactions

Final answer:

No reaction is expected when (NH4)2S(aq) and SrCl2(aq) are mixed, as solubility rules suggest no insoluble salts will form, leading to NOREACTION.

Explanation:

When (NH4)2S(aq) and SrCl2(aq) are mixed together, we expect a reaction where the cations (NH4+ and Sr2+) and anions (S2- and Cl-) exchange partners if any of them can form an insoluble salt. Looking at solubility rules, we know that most sulfides are insoluble except those of alkali metals and ammonium, and most chlorides are soluble except for Ag+, Pb2+, and Hg22+. Given that neither NH4+ nor Sr2+ forms an insoluble chloride and SrS is not listed as an insoluble sulfide, we can predict that no visible reaction will occur when these solutions are mixed. Therefore, the chemical equation to represent this mixture is NOREACTION.

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1. An isotope of cesium-137 has a half-life of 30 years. If 5.0 g of cesium-137 decays over 60 years, how many grams will remain?

Answers

Answer:

1.25 g

Explanation:

Now we have to use the formula;

N/No = (1/2)^t/t1/2

N= mass of cesium-137 left after a time t (the unknown)

No= mass of cesium-137 present at the beginning = 5.0 g

t= time taken for 5.0 g of cesium-137 to decay =60 years

t1/2= half life of cesium-137= 30 years

Substituting values;

N/5= (1/2)^60/30

N/5= (1/2)^2

N/5= 1/4

4N= 5

N= 5/4

N= 1.25 g

Therefore, 1.25 g of cesium-137 will remain after 60 years.

Calculate the pH of the solution formed when 45.0 mL of 0.100M NaOH solution is added to 50.0 mL of 0.100M CH3COOH (Ka for acetic acid = 1.8 x10-5 ).

Answers

The study of chemicals and bonds is called chemistry. There are two types of elements are there and these rare metals and nonmetals.

The correct answer is 5.59.

What is PH?

pH, historically denoting "potential of hydrogen" is a scale used to specify the acidity or basicity of an aqueous solution.
Acidic solutions are measured to have lower pH values than basic or alkaline solutions

The pH of the solution will be due to excessive acid left and the salt formed. Thus, it will form a buffer solution.

The pH of buffer solution is calculated from Henderson Hassalbalch's equation, which is:

$pH= pka+log(salt)/(acid)$

$pka= -log(ka)$

$pka =-log(1.8*10^(-5)$

The moles of acid are taken as:-

$moles = M*V$

$0.5*50=5$

The moles of the base are taken as:-

$moles = M*V$

$0.1*45=4.5$

moles of acid left is 0.5

Place all the values to the equation:-

$pH=4.74+log(4.5)/(0.5) \n=5.69$

Hence, the correct answer is 5.69.

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Answer:

pH of soltion will be 5.69

Explanation:

The pH of the solution will be due to excessive acid left and the salt formed. Thus, it will form a buffer solution.

The pH of buffer solution is calculated from Henderson Hassalbalch's equation, which is:

$pH=pKa+log(([salt])/([acid]) )$

$pKa=-log(Ka)$

$pKa=-log(1.8X10^(-5))=4.74$

The moles of acid taken :

$moles=molarityXvolume=0.1X50=5mmol$

The moles of base taken:

$moles=molarityXvolume=0.1X45=4.5mmol$

The moles of acid left after reaction :

$5-4.5=05mmol$

The moles of salt formed = 4.5mmol

Putting values in equation

$pH=pKa+log(([salt])/([acid]) )=4.74+log((4.5)/(0.5))=5.69$

Which chemical equation follows the law of conservation of mass?

Answers

The chemical equation presented in option A follows the law of conservation of mass.

The principle of conservation of mass states, mass can neither be created nor destroyed but can be transformed from one form to another.

A reaction that follows the law of conservation of mass, must have equal number of moles each elements in reactants side and products side.

Only option A follows the law of conservation of mass;

$2LiOH \ + \ + H_2CO_3 \ ---> \ Li_2CO_3 \ + \ 2H_2O$

Thus, we can conclude that the chemical equation presented in option A follows the law of conservation of mass.

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Answer:

Option A

Explanation:

The expression that obeys the law of conservation of mass is choice A;

2LiOH + H₂CO₃ → Li₂CO₃ + 2H₂O

According to the law of conservation of mass; "in a chemical reaction, matter is neither created nor destroyed". By this law, mass is usually conserved.

The equation shows that mass is conserved because the number of moles of each specie is found on both sides

Number of moles

Li O H C

Reactants 2 5 4 1

Products 2 5 4 1

This shows that mass is indeed conserved.

An organic compound was extracted into dichloromethane and then the aqueous layer is shaken with saturated sodium chloride solution. What is the purpose of the sodium chloride? A. to decrease the solubility of the organic product in water B. to raise the density of the aqueous layer so that it will be the bottom layer C. to lower the boiling point of the water to cause the water D. to be immiscible with the methylene chloride

Answers

Answer: option A. to decrease the solubility of the organic product in water

Explanation: sodium chloride solution act as a drying agent to remove water from an organic compound that is in solution. The salt water works to pull the water from the organic layer to the water layer,therby decreasing

The value of the Solubility Product Constant for lead phosphate is ____________Write the reaction that corresponds to this Ksp value.

_______(Aq,S,L) +_______(Aq,S,L) <-------->_______(Aq,S,L) +_______(Aq,S,L)

Ksp values are found by clicking on the "Tables" link.

Use the pull-down menus to specify the state of each reactant or product.

If a box is not needed leave it blank.

Answers

Answer: The reaction for the $K_(sp)$ value of lead phosphate is given below and the value of solubility product for the same is $3.0\rightarrow 10^(-44)$

Explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio. It is expressed as $K_(sp)$

The chemical formula of lead phosphate is $Pb_3(PO_4)_2$

The equation for the hydration of the lead phosphate is given as:

$Pb_3(PO_4)_2(s)+H_2O(l)\rightarrow 3Pb^(2+)(aq.)+2PO_4^(3-)(aq.)$

The solubility product of lead phosphate is $3.0\rightarrow 10^(-44)$ . This means that it is highly insoluble in water as the solubility product is very very low.

Hence, the reaction for the $K_(sp)$ value of lead phosphate is given above and the value of solubility product for the same is $3.0\rightarrow 10^(-44)$

Other Questions

Calculate the molar mass of each compound given below. c4h4

A solution of NaOH has a concentration of 25.00% by mass. What mass of NaOH is present in 0.250 g of this solution? Use the periodic table in the toolbar if needed.

A weather system moving through the American Midwest produced rain with an average pH of 5.02. By the time the system reached New England, the rain it produced had an average pH of 4.66. How much more acidic was the rain falling in New England?

The higher the pH, the less acidic the solution

Answers

Final answer:

Explanation:

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Related Questions

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Further Explanation

Chemical equation

Types of chemical reactions

Precipitation reaction

Displacement reaction

Decomposition reaction

Neutralization reaction

Redox reactions

Learn more about:

Final answer:

Explanation:

Learn more about Chemical Reaction Prediction here:

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What is PH?

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