The solubility product of Car 3.4×10-11 What is its solubility in 0.01 M solution of NaF

a. 3.4×10-7 mol L-1

C. 3.4×10-2 mol L-1

b. 3.4×10-5 mol L-1

d. 3.4 mol L-1

To solve this problem, we'll assume that the compound in question is \( CaF_2 \), which dissociates in water as follows:

\[
CaF_2 \leftrightarrow Ca^{2+} + 2F^-
\]

Given:
- The solubility product, \( K_{sp} = 3.4 \times 10^{-11} \)
- The concentration of \( NaF \) is 0.01 M, meaning the initial concentration of \( F^- \) ions is 0.01 M.

Let \( S \) be the solubility of \( CaF_2 \) in the 0.01 M \( NaF \) solution. The concentration of \( Ca^{2+} \) ions will be \( S \) and the concentration of \( F^- \) ions will be \( 0.01 + 2S \) (since each \( CaF_2 \) produces 2 \( F^- \) ions).

The solubility product expression for \( CaF_2 \) is:

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

Substituting the values:

\[
3.4 \times 10^{-11} = S \times (0.01 + 2S)^2
\]

Since \( K_{sp} \) is very small, the solubility \( S \) will be much smaller than 0.01 M, so \( 2S \) can be neglected compared to 0.01 M:

\[
3.4 \times 10^{-11} \approx S \times (0.01)^2
\]

\[
3.4 \times 10^{-11} = S \times 10^{-4}
\]

\[
S = \frac{3.4 \times 10^{-11}}{10^{-4}} = 3.4 \times 10^{-7} \text{ mol/L}
\]

Thus, the correct answer is:

**a. \( 3.4 \times 10^{-7} \text{ mol/L} \)**

Would you like more details or have any questions? Here are some related questions you might find interesting:

1. How does the common ion effect influence the solubility of \( CaF_2 \)?
2. What would be the solubility of \( CaF_2 \) in pure water?
3. How does \( K_{sp} \) vary with temperature?
4. What happens to the solubility of \( CaF_2 \) if \( NaF \) concentration increases?
5. Can you explain the significance of the solubility product constant?

**Tip:** When dealing with solubility in the presence of a common ion, always consider simplifying assumptions to make the calculation more straightforward.

Learn how to calculate the solubility of CaF2 in a 0.01 M NaF solution using the solubility product constant. Understand the step-by-step solution and apply the solubility product expression. Suitable for high school students studying chemical equilibrium.

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