Question at position 25 SrCO3 has a Ksp of 25.0 x 10–10 at a certain temperature. What is its molar solubility at this temperature? SrCO3 has a Ksp of 25.0 x 10–10 at a certain temperature. What is its molar solubility at this temperature? 1.0 x 10–3 M 5.0 x 10–5 M 2.0 x 10–5 M 1.0 x 10–4 MClear my selection

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BlackTom AI · Accepted Answer

The question asks for the molar solubility of SrCO3 given its Ksp at a certain temperature. 
First, write the dissolution equilibrium: SrCO3(s) ⇌ Sr2+(aq) + CO3^2−(aq). The solubility product expression is Ksp = [Sr2+][CO3^2−]. If the molar solubility is s, then both ions are produced in a 1:1 ratio, so [Sr2+] = s and [CO3^2−] = s. Therefore Ksp = s × s = s^2. 
Next, substitute the given Ksp value. The problem states Ksp = 25.0 × 10^−10, which is 2.50 × 10^−9 when written in standard scientific notation. So s^2 = 2.50 × 10^−9. Taking the square root yields s = √(2.50 × 10^−9).
Now estimate the square root. √2.50 ≈ 1.58, and √(10^−9) = 10^−4.5 ≈ 3.16 × 10^−5. Multiplying gives s ≈ 1.58 × 3.16 × 10^−5 ≈ 5.0 × 10^−5 M. 
With this framework, evaluate each option:
- 1.0 × 10^−3 M: This value would imply Ksp ≈ (1.0 × 10^−3)^2 = 1.0 × 10^−6, which is far larger than 2.50 × 10^−9, so this cannot be correct.
- 5.0 × 10^−5 M: If s = 5.0 × 10^−5, then Ksp = (5.0 × 10^−5)^2 = 25 × 10^−10 = 2.5 × 10^−9, exactly matching the given Ksp. This is consistent and correct.
- 2.0 × 10^−5 M: Here, Ksp would be (2.0 × 10^−5)^2 = 4.0 × 10^−10, which is smaller than the provided Ksp, so this solubility is too low.
- 1.0 × 10^−4 M: This yields Ksp = (1.0 × 10^−4)^2 = 1.0 × 10^−8, which is much larger than the given value, so it cannot be correct.

In summary, only the second option yields a Ksp that matches the given value, while the others produce either much larger or much smaller Ksp values. The reasoning relies on the simple 1:1 dissolution stoichiometry of SrCO3 and the direct application of the Ksp expression.

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