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Andrews [41]
3 years ago
10

In the pictured reaction, NH4, would be acting as the

Chemistry
1 answer:
Svetach [21]3 years ago
7 0

Answer:

conjugate acid

Explanation:

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What is the solubility (in g/L) of calcium fluoride at 25°C? The solubility product constant for calcium fluoride is 3.4 × 10–11
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<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

CaF_2\rightleftharpoons Ca^{2+}+2F^-

                s       2s

The expression for solubility constant for this reaction will be:

K_{sp}=[Ca^{2+}][F^-]^2

We are given:

K_{sp}=3.4\times 10^{-11}

Putting values in above equation, we get:

3.4\times 10^{-11}=(s)\times (2s)^2\\\\3.4\times 10^{-11}=4s^3\\\\s=2.04\times 10^{-4}mol/L

To calculate the solubility in g/L, we will multiply the calculated solubility with the molar mass of calcium fluoride:

Molar mass of calcium fluoride = 78 g/mol

Multiplying the solubility product, we get:

s=2.04\times 10^{-4}mol/L\times 78g/mol=159.12\times 10^{-4}g/L=0.015g/L

Hence, the correct answer is Option b.

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