Question

Question 3 out of 1 points The Ksp of Al(OH)3 is 1.0x 10. What is the solubility of AI(OH)3 in 0.0000010 M AI(NO3)3? Give your answer using scientific notation and to 2 significant figuresi.e., one decimal place). [Blank1] Specified Answer for: Blank1 1.510-11 Correct Answers for: Blank1 Evaluation Method Correct Answer Case Sensitivity Exact Match 3.3e-10

Answer 1

Molar concentrations of Al (NO_3)_3 = 0.0000010 M Equation for the dissociation of Al (NO_3)_3: Al (NO_3)3 rightarrow AI^3 + + 3NO_3^- 1 mole of Al (NO_3)_3 forms 1 mole of Al^3 +, [Al^3 +] = [Al (NO_3)_3] = 0.0000010 M Solubility product K_sp = [Al^3 +] [OH^-]^3 1.0 times 10^-33 = (0.0000010 + s) (3s)^3 Since the solubility product is too low, we can assume that 0.0000010 + s equivalent 0.0000010 1.0 times 10^-33 = (0.0000010 + s) (3s)^3 1.0 times 10^-33 = (0.0000010) 27s^3 27 s^3 = 1.0 times 10^-33/(0.0000010) s^3 = 1.0 times 10^-33/27 (0.0000010) s = 3 squareroot 1.0 times 10^-33/27 (0.0000010) = 3 squareroot 0.037 times 10^-27 = 0.33 times 10^-9 M = 3.3 times 10^-10 M Thus, molar solubility of Al (OH)_3 in 0.0000010 M Al (NO_3)_3 is 3.3 times 10^-10 M