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Alexxandr [17]
3 years ago
14

since Ag and Cl are in equilibrium with AgCl, find Ksp for agcl from the concentrations. Write the expression for ksp AgCl and d

etermine its value
Chemistry
1 answer:
melamori03 [73]3 years ago
6 0

Answer:

1.69 ×10^-10

Explanation:

Given that the equation for the dissolution of AgCl in water is;

AgCl(s) ⇄Ag^+(aq) + Cl^+(aq)

Also, silver ion and chloride ion are in equilibrium with the undissociated AgCl hence we can write;

Ksp= [Ag+][Cl-]/ [AgCl]

Since the activity of the pure sold is 1, we now have;Ksp= [Ag+][Cl-]

If we know the solubility of AgCl in pure water to be 1.3 x 10^-5 M, from standard tables, and [Ag+]=[Cl-]= 1.3 x 10^-5 M = x

Then;

Ksp= x^2

Ksp= (1.3 x 10^-5)^2

Ksp= 1.69 ×10^-10

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1) Compund Ir (x) O(y)

2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g

3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g

4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g

5) Convert grams to moles

moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles

moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131

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Divide by the least of the number of moles, i.e. 0.00656

Ir: 0.00656 / 0.00656 = 1

O: 0.0131 / 0.00656 = 2

=> Empirical formula = Ir O2 (where 2 is the superscript for O)

Answer: Ir O2
5 0
3 years ago
Read 2 more answers
A solution contains 0.036 M Cu2+ and 0.044 M Fe2+. A solution containing sulfide ions is added to selectively precipitate one of
Ratling [72]

Answer:

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

Explanation:

<u>Step 1: </u>Data given

The solution contains 0.036 M Cu2+ and 0.044 M Fe2+

Ksp (CuS) = 1.3 × 10-36

Ksp (FeS) = 6.3 × 10-18

Step 2:  Calculate precipitate

CuS → Cu^2+ + S^2-         Ksp= 1.3*10^-36

FeS → Fe^2+ + S^2-      Ksp= 6.3*10^-18

Calculate the minimum of amount needed to form precipitates:

Q=Ksp

<u>For copper</u>  we have:  Ksp=[Cu2+]*[S2-]

Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]

[S2-]= 3.61*10^-35 M

<u>For Iron</u>  we have: Ksp=[Fe2+]*[S2-]

Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]

[S2-]= 1.43*10^-16 M

CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

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Qsp = 6.1 * 10^-8

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