The Kp for praseodymium (III) hydroxide, Pr(OH)3, is 3.39 x 10-24. What pH, to two places past the decimal, will just precipitate out praseodymium (III) hydroxide in a 0.659 M solution of Pr(NO3);?
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![**Question for Students: Calculating pH for Precipitation of Praseodymium (III) Hydroxide**
The solubility product constant (\(K_{sp}\)) for praseodymium (III) hydroxide, \( \text{Pr(OH)}_3 \), is \( 3.39 \times 10^{-24} \).
**Problem Statement:**
What pH, to two decimal places, will just precipitate out praseodymium (III) hydroxide in a 0.659 M solution of \( \text{Pr(NO}_3)_3 \)?
**Explanation:**
To determine this, students should recall the relationship between the solubility product constant and the concentrations of ions at equilibrium. They need to solve for the hydroxide ion concentration ([OH⁻]) and then use this to find the corresponding pH.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdb02c8bf-339c-4ae2-9d8b-8db604f1f70e%2F11bc4860-8fca-4466-b8b4-e21bc39d2cb0%2Frciiudo_processed.jpeg&w=3840&q=75)
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