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yan [13]
4 years ago
13

A solution contains one or more of the following ions: Ag+, Ca2+, and Fe2+. When sodium chloride is added to the solution, no pr

ecipitate forms. When sodium sulfate is added to the solution, a white precipitate forms. Then, the precipitate is filtered off and sodium carbonate is added to the remaining solution, and a precipitate forms. Which of the ions were present in the original solution? Select all that apply.
Chemistry
2 answers:
Vsevolod [243]4 years ago
7 0

Answer:

Ca2+ and Fe2+

Explanation:

When a solution is added to the original one, a single replacement reaction will occur. So, the cation will be substituted for the cation presented in the solution. If the new compound is soluble, it will dissociate, if not, it will precipitate.

First, sodium chloride is added, so, the sodium cation will be substituted. The possibilities of new formations are AgCl, CaCl2, and FeCl2. Both CaCl2 and FeCl2 are solubles, but AgCl is insoluble, thus, if it was present, it would form a precipitate, so Ag+ is not present.

When sodium sulfate is added, a precipitate is formed. The sodium can be replaced, forming CaSO4 or FeSO4. CaSO4 is insoluble, so it forms a precipitate, which may be the solid formed. FeSO4 is soluble.

Then, sodium carbonate is added and can form FeCO3, which is insoluble and can be the solid formed. So, possible ions are Ca2+ and Fe2+.

Ksju [112]4 years ago
6 0

Answer:

the ion present in the original solution is Ca2+

Explanation:

Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a precipitate.

<u>Step1</u> : If we add Nacl to the solution, there is no precipitate formed

⇒The only possible ion that can form a precipate with Cl- is Ag+; since there is no precipitate formed, Ag+ is not present

<u>Step2</u> : If we add Na2SO4 to the solution, a white precipitate is formed

The possible ions to bind at SO42- are Ca2+ and Fe2+

But the white precipitate formed, points in the direction of Ca2+

⇒This means calcium is present

<u>Step3</u> : If we add Na2CO3 to the filtered solution, there is a precipate formed

Ca2+ will bind also with CO32- and form a precipitate

So the ion present in the original solution is Ca2+

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The complete table is inserted.

A table is given,

Formulas used:

pH=  -log(H⁺)

pOH=  -log(OH⁻)

pH+ pOH=14

Calculations:

For A: (H⁺)=2×10⁻⁸M

Using the pH formula:

pH=  -log(H⁺)=-log(2×10⁻⁸)=7.69

pOH=14 - 7.69=6.3

Calculating OH concentration,

pOH=  -log(OH⁻)

6.3= -log(OH⁻)

(OH⁻)=5.011×10⁻⁷M

Hence, the nature of A is basic.

Similarily,

For B,

(OH⁻)=1×10⁻⁷

Using the pH formula:

pOH=  -log(OH⁻)= -log(1×10⁻⁷)=7

pH=14-7=7

Calculating H concentration,

pH=  -log(H⁺)

7= -log(H⁺)

(H⁺)=1×10⁻⁷M

Hence, the nature of B is neutral.

Similarily,

For C,

pH=12.3

Using the pH formula:

pOH=14-12.3=1.7

Calculating H concentration,

pH=  -log(H⁺)

12.3= -log(H⁺)

(H⁺)=5.011×10⁻¹³M

Calculating OH concentration,

pOH=  -log(OH⁻)

1.7= -log(OH⁻)

(OH⁻)=1.99×10⁻²M

Hence, the nature of C is Basic.

Similarily,

For D,

pOH=6.8

Using the pH formula:

pH=14-6.8=7.2

Calculating H concentration,

pH=  -log(H⁺)

7.2= -log(H⁺)

(H⁺)=6.309×10⁻⁸M

Calculating OH concentration,

pOH=  -log(OH⁻)

6.8= -log(OH⁻)

(OH⁻)=1.58×10⁻⁷M

Hence, the nature of D is basic.

Learn more about the acid and bases here:

brainly.com/question/16189013

#SPJ10

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