The solubility product constant of PbI2(s) is 7.1 10-9. How many ...

[Starlight]

Question 1

The solubility product constant of Mg(OH)₂ is 9.0 × 10^-12. If an aqueous solution is 0.010 M with respect to Mg²⁺ ion, the amount of [OH^-] required to start the precipitation of Mg(OH)₂(s) is:
◦ 1.5 × 10^-7 M
◦ 3.0 × 10^-5 M
◦ 3.0 × 10^-7 M
◦ 9.0 × 10^-10 M
◦ 1.5 × 10^-5 M

Question 2

The solubility product constant of PbI₂(s) is 7.1 × 10^-9. How many moles of PbI₂ will precipitate if 250 mL of a 0.200 M solution of NaI(aq) are added to 150 mL of 0.100 M solution of Pb(NO₃)₂(aq)? You may neglect hydrolysis.
◦ 0.050 mol
◦ 1.3 × 10^-5 mol
◦ 0.015 mol
◦ 5.6 × 10^-3 mol
◦ 0.040 mol

[Chocorrol77]

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[Starlight]

Great answer, keep it coming

[carlsona147]

Thanks for the timely response, appreciate it