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Author Question: The solubility product constant of PbI2(s) is 7.1 10-9. How many moles of PbI2 will precipitate if ... (Read 63 times)

Starlight

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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


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Marked as best answer by Starlight on Jul 8, 2021

Chocorrol77

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Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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Starlight

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Reply 2 on: Jul 8, 2021
Thanks for the timely response, appreciate it


blakcmamba

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Reply 3 on: Yesterday
Gracias!

 

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