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Flauer [41]
3 years ago
11

Calculate the solubility of CuX (Ksp=[Cu2+][X2−]=1.27×10−36) in a solution that is 0.200 M in NaCN.

Chemistry
2 answers:
Advocard [28]3 years ago
5 0

Answer:

The solubility of CuX is 1.425x10⁻⁷M

Explanation:

Given:

initial concentration of NaCN=0.2M

Ksp=1.27x10⁻³⁶

The reaction are:

CuX → Cu²⁺ + X²⁺, Ksp=1.27x10⁻³⁶

Cu²⁺ + 4CN⁻ → (Cu(CN)₄)²⁻, Kf=1x10²⁵

The overall reaction is:

CuX + 4CN⁻ → (Cu(CN)₄)²⁻ + X²⁺

The equilibrium constant is:

K=Ksp*Kf=1.27x10⁻³⁶*1x10²⁵=1.27x10⁻¹¹

              CuX      +        4CN⁻      →       (Cu(CN)₄)²⁻ +        X²⁺

I              -                      0.2                    0                         0

C            -                       -4                     +x                        +x

E             -                      0.2-4                x                          x

The equation for equilibrium is:

K=\frac{[Cu(CN)4]^{2} [X]}{[CN]^{4} } \\1.27x10^{-11} =\frac{x^{2} }{(0.2-x)^{4} }

Here, solving for x:

x=1.425x10⁻⁷M=CuX

Tomtit [17]3 years ago
3 0

Answer:

Solubility= 1.08×10-12

Explanation:

Take the cube root of 1.27×10-36

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1. For each of the following, convert the word equation into a formula equation, BUT do not balance! (4 pts each = 12 pts)
daser333 [38]

Answer:

  • 1a) BaClO₃(s) → BaCl₂(g) + O₂(g)

  • 1b) Cl₂(g) + K₃N(s) → N₂(g) + KCl(s)

  • 1c) Na₃N(aq) + Al(BrO₃)₃(aq) → AlN(s) + Na(BrO₃)₃(aq)

  • 2a) Calcium hydroxide and hydrogen gas

  • 2b) Tin(II) silicate and Lead(IV) permanganate

  • 2c) Magnesium oxide and water

  • 2d) No product

  • 2e) Mercury and iodine

  • 2f) Calcium chloride and iodine

  • 2g) Strontium phosphite and cesium nitride

  • 2h) Carbon dioxide, water, and sulfur dioxide

  • 2i) Iron oxide(III) and carbon dioxide

  • 2j) Magnesium acetate and hydrogen gas

  • 2k) Calcium iodide

Explanation:

1. For each of the following, convert the word equation into a formula equation, BUT do not balance!

a) Barium chlorate → Barium chloride + Oxygen

<u>1. Chemical formulas</u>

Barium chlorate:

  • It is a salt: an ionic compound.
  • Barium has oxidation state +2
  • Chlorate is the ion ClO₃⁻
  • Swap the oxidation numbers to write the subscripts: 2 goes to ClO₃ and 1 goes to Ba
  • Chemical formula Ba(ClO₃)₂
  • It is solid: Ba(ClO₃)₂(s)

Barium chloride:

  • It is a salt: an ionic compount
  • Barium has oxidation state +2
  • Chlorine is in oxidation state -1
  • Swap the numbers to write the subscripts: 2 goes to Cl and 1 goes to Ba
  • BaCl₂
  • It is solid BaCl₂(s)

Oxygen:

  • It is a diatomic gas molecule
  • O₂(g)

<u />

<u>2. Write the unbalanced molecular equation:</u>

  • BaClO₃(s) → BaCl₂(s) + O₂(g)

b) Chlorine + Potassium nitride → Nitrogen + Potassium chloride

<u>1. Chemical formulas</u>

Chlorine:

  • It is a diatomic gas molecule
  • Cl₂(g)

Potassum nitride

  • It is a salt
  • Potassium has oxidation state +1
  • Nitrogen is with oxidation state +3
  • Swap the oxidation states
  • K₃N
  • It is solid: K₃N(s)

Nitrogen:

  • It is a diatomic gas
  • N₂(g)

Potassium chloride

  • It is a salt (ionic compound)
  • Potassium has oxidation state +1
  • Chlorine is in oxidation state -1
  • Swap the oxidation numbers
  • KCl
  • It is solid: KCl(s)

<u>2. Write the unbalanced molecular equation</u>

<u />

  • Cl₂(g) + K₃N(s) → N₂(g) + KCl(s)

c) Sodium nitride + Aluminum bromate → Aluminum nitride + Sodium bromate

<u>1. Chemical formulas</u>

Sodium nitride

  • It is a salt (ionic compound)
  • Sodium has oxidation state +1
  • Nitrogen is with oxidation state -3
  • Swap the oxidation numbers
  • Na₃N
  • It is in aqueous solution
  • Na₃N (aq)

Aluminum bromate

  • Salt
  • Aluminum has oxidation state +3
  • Bromate is the ion BrO₃⁻
  • Swap the oxidation states
  • Al(BrO₃)₃ (aq)

Aluminum nitride

  • Both Al and N have oxidation state 3, which simply
  • AlN(s). It is not soluble in water.

Sodium bromate

  • Na(BrO₃)₃ (aq)

<u>2. Write the unbalanced molecular equation</u>

  • Na₃N(aq) + Al(BrO₃)₃(aq) → AlN(s) + Na(BrO₃)₃(aq)

<h2>This is a long answer with more than 5,000 charaters; thus, I have to add the rest of the explanations on a separate file.</h2><h2></h2><h2>The attached file contains the complete answer.</h2>
Download pdf
4 0
3 years ago
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