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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 64 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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Starlight

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Reply 2 on: Jul 8, 2021
Gracias!


lcapri7

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Reply 3 on: Yesterday
:D TYSM

 

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