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3 years ago

What is the molar solubility of MgF2 in a 0.36 M Mg(NO3)2 solution? For MgF2, Ksp = 8.4 × 10^–8

Chemistry

1 answer:

Valentin [98]3 years ago

7 0

Answer:

2.4 × 10⁻⁴ M

Explanation:

Step 1: Calculate the concentration of Mg²⁺ coming from Mg(NO₃)₂

Mg(NO₃)₂ is a strong electrolyte and the molar ratio of Mg(NO₃)₂ to Mg²⁺ is 1:1. The initial molar concentration of Mg²⁺ is 1/1 × 0.36 M = 0.36 M.

Step 2: Make an ICE chart for the solution of MgF₂

MgF₂(s) ⇄ Mg²⁺(aq) + 2 F⁻(aq)

I 0.36 0

C +S +2S

E 0.36+S 2S

The solubility product constant is:

Ksp = [Mg²⁺] × [F⁻]² = (0.36+S) × (2S)²

Since S <<< 0.36, 0.36+S ≈ 0.36.

Ksp = 0.36 × 4S² = 8.4 × 10⁻⁸

S = 2.4 × 10⁻⁴ M

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3 years ago

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4 years ago

When you calculate the Ksp of calcium fluoride, CaF2, and the concentration of the F- ion is 4.2 x 10-4M, what is the concentrat

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Calculate its molar solubility in a saturated aqueous solution that is also 0.050 M in fluoride ion, F-. Ksp = (x)(2x + 0.05)2. 2x = 4.210 4

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2 years ago

A method that uses a mathematical equation and the half-life of decay to determine the exact date of a fossil is known as ______

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3 years ago

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For the reaction: CH3NH2(aq) + H2O(aq) ⇌ CH3NH3 +(aq) + OH- Determine the change in the pH (ΔpH) for the addition of 6.7 M CH3NH

Korolek [52]

Answer:

The change in the pH (ΔpH) is 2,17

Explanation:

The reaction:

CH₃NH₂(aq) + H₂O(aq) ⇌ CH₃NH₃⁺(aq) + OH⁻

$kb = \frac{[OH^{-}][CH_{3}NH_{3}^+]}{[CH_{3}NH_{2}]}$ (1)

In equilibrium, a solution of CH₃NH₂ 4,7M produces:

[CH₃NH₂] = 4,7 - x

[CH₃NH₃⁺] = x

[OH⁻] = x

Replacing in (1):

$4,38x10^{-4} = \frac{x^2}{4,7-x}$

x² + 4,38x10⁻⁴x - 2,0586x10⁻³ = 0

The solutions are:

x = -0,0456 No physical sense. There are not negative concentrations.

x = 0,04515 Real answer.

The concentration of [OH⁻] is 0,04515 M.

As pOH = -log [OH⁻] And pH+pOH = 14. The pH of this solution is:

pH = 12,65

The addition of 6,7M produce this changes in concentrations:

[CH₃NH₂] = 4,656 + x

[CH₃NH₃⁺] = 6,74515 - x

[OH⁻] = 0,04515 - x

Replacing in (1) you will obtain:

x² - 6,7907x + 0,3025 = 0

Solving for x:

x = 6,74586 No physical sense

x = 0,04484 Real answer.

Thus, [OH⁻] = 0,04515 - 0,044842 = 3,08x10⁻⁴M

pOH = 3,51.

pH = 10,49

Thus ΔpH is 12,65 - 10,49 = 2,16 ≈ 2,17

I hope it helps!

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4 years ago