What is the solubility of Ag2SO4 if the Ksp of Ag2SO4 is 1.4 × 10^-5?
A. 1.5 × 10-2 M
B. 2.4 × 10-2 M
C. 7.4 × 10-3 M
D. 3.7 × 10-3 M
Follow the instructions for part a which I worked for another student.
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https://www.jiskha.com/display.cgi?id=1322854986
To find the solubility of Ag2SO4, we need to calculate the concentration of Ag+ ions in a saturated solution using the solubility product constant (Ksp) for Ag2SO4.
Ag2SO4 dissociates in water to form two Ag+ ions and one SO4 2- ion. Therefore, the balanced equation for the dissociation of Ag2SO4 is:
Ag2SO4(s) ⇌ 2Ag+(aq) + SO4 2-(aq)
The Ksp expression for Ag2SO4 is:
Ksp = [Ag+]^2 * [SO4 2-]
Given that the Ksp of Ag2SO4 is 1.4 × 10^-5, we can set up the equation as follows:
1.4 × 10^-5 = (2x)^2 * (x)
where x represents the molar solubility of Ag2SO4 and 2x represents the concentration of Ag+ ions.
Simplifying the equation:
1.4 × 10^-5 = 4x^3
To solve for x, take the cube root of both sides:
x = (1.4 × 10^-5)^(1/3)
Using a calculator, we find:
x ≈ 7.4 × 10^-3 M
Therefore, the solubility of Ag2SO4 is approximately 7.4 × 10^-3 M.
Therefore, the correct answer is C. 7.4 × 10^-3 M.