Solubility Homework Set Worked Out Solutions

Whatisthephofasaturatedsolutionofcuoh2ksp261019

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Unformatted text preview: (aq) BiI3(s) ___ Initial ‐‐‐‐‐ 0.20 M 0 M Change ‐‐‐‐‐ +1x M +3x M Equilibrium ‐‐‐‐‐ x+ 0.20 M 3x M Ksp=[ Bi3+] [I‐]3= (x+0.20)(3x)3=(0.20)(27x3)=8.1 x 10‐19 6 Dr. Fus CHEM 123 x= 5.3 x 10‐7 M 28. As the pH decreases, how will the solubility of Cu(OH)2 be affected? To determine the effect of an acid/base or change in pH, write out the solubility equilibrium expression: Cu2+(aq) + 2OH‐(aq) Cu(OH)2 (s) If the pH is decreased, the concentration of H+ ions increases. These H+ ions can neutralize the OH‐ ions from the above equilibrium, forming H2O, and this decreases the concentration of OH‐ in solution. According to LeChatelier’s Principle, the equilibrium will then shift to the right, thereby increasing the solubility of Cu(OH)2(s). 29. The Ksp for Zn(OH)2 is 5.0 10‐17. Determine the molar solubility of Zn(OH)2 in buffered solution with a pH of 11.50? First, determine the concentration of OH‐ when the pH is 11.50: pH=‐log[H+]=11.50 [H+]=3.16 x 10‐12 M +][OH‐]=1.0 x 10‐14 (always true) [H (3.16 x 10‐12 M) +][OH‐]=1.0 x 10‐14 [OH‐]=3.16 x 10‐3 M Now, the problem is similar to a common‐ion problem. However, because the solution is buffered, it resists changes in pH. Therefore, the concentration of OH­ does NOT change: Zn2+(aq) + 2OH‐(aq) Zn(OH)2(s) ___ Initial ‐‐‐‐‐ 0 M 3.16 x 10­3 M Change ‐‐‐‐‐ +1x M NO CHANGE Equilibrium ‐‐‐‐‐ x M 3.16 x 10­3 M Ksp=[ Zn2+] [OH‐]2=(x)( 3.16 x 10‐3)2=5.0 x 10‐17 x= 5.0 x 10‐12 M 30. Calculate the solubility of Cu(OH)2 in a solution buffered at pH = 8.50. Ksp for Cu(OH)2 = 1.6 x 10‐19 Initial Change Equilibrium pH=‐log[H+]=8.50 [H+]=3.16 x 10‐9 M [H+][OH‐]=1.0 x 10‐14 (always true) (3.16 x 10‐9 M) +][OH‐]=1.0 x 10‐14 [OH‐]=3.16 x 10‐6 M Cu2+(aq) + 2OH‐(aq) Cu(OH)2(s) ___ ‐‐‐‐‐ 0 M 3.16 x 10­6 M ‐‐‐‐‐ +1x M NO CHANGE ‐‐‐‐‐ x M 3.16 x 10­6 M Ksp=[ Cu2+] [OH‐]2=(x)( 3.16 x 10‐6)2=1.6 x 10‐19 x= 1.6 x 10‐8 M 7 Dr. Fus CHEM 123 31. Calculate the molar solubility of Mn(OH)2 when buffered at pH = 11.40. The Ksp for Mn(OH)2 is 1.6 x 10­13. Initial Change Equilibrium pH=‐log[H+]=11.40 [H+]=3.98 x 10‐12 M [H+][OH‐]=1.0 x 10‐14 (always true) (3.98 x 10‐12 M) +][OH‐]=1.0 x 10‐14 [OH‐]=2.51 x 10‐3 M Mn2+(aq) + 2OH‐(aq) Mn(OH)2(s) _ ‐‐‐‐‐ 0 M 2.51 x 10­3 M ‐‐‐‐‐ +1x M NO CHANGE ‐‐‐‐‐ x M 2.51 x 10­3 M 2+] [OH‐]2=(x)( 2.51 x 10‐3)2=1.6 x 10‐13 Ksp=[ Mn x= 2.5 x 10‐8 M 32. What is the pH of a saturated solution of Cu(OH)2 (Ksp = 2.6 10‐19)? To determine the pH, we need to determine the concentration of H+ using an ICE table: Cu2+(aq) + 2OH‐(aq) Cu(OH)2(s) ___ Initial ‐‐‐‐‐ 0 M 0 M Change ‐‐‐‐‐ +1x M +2x M Equilibrium ‐‐‐‐‐ x M 2x M Ksp=[Cu2+][OH‐]2=(x)( 2x)2=4x3=2.6 x 10‐19 x= 4.0 x 10‐7 M *Remember to multiply x by 2 in this problem because the equilibrium concentration of...
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This homework help was uploaded on 03/28/2014 for the course CHEM 123 taught by Professor Larosa during the Spring '06 term at Ohio State.

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