Solid sodium bromide is slowly added to a solution that is 0.010 M in Cu+ ions a

Question

Solid sodium bromide is slowly added to a solution that is 0.010 M in Cu+ ions and 0.010 M Ag+ ions. A) which compound precipitates first? B) if silver bromide precipitates first, what is the silver ion concentration when copper (I) bromide begins to precipitate? C) what percent of silver ions remain in solution when the copper (I) bromide begins to precipitate? Solid sodium bromide is slowly added to a solution that is 0.010 M in Cu+ ions and 0.010 M Ag+ ions. A) which compound precipitates first? B) if silver bromide precipitates first, what is the silver ion concentration when copper (I) bromide begins to precipitate? C) what percent of silver ions remain in solution when the copper (I) bromide begins to precipitate? A) which compound precipitates first? B) if silver bromide precipitates first, what is the silver ion concentration when copper (I) bromide begins to precipitate? C) what percent of silver ions remain in solution when the copper (I) bromide begins to precipitate?

Explanation / Answer

Cu(Br) = 6.27×10-9

AgBr = 5.35×10-13

Since Ksp of Ag is smaller, expect Ag+ to precipitate first

b)

Ksp = Cu*Br

6.27×10-9 = [Cu][Br]

solve for Br

Br = (6.27*10^-9)/(0.01) = 6.27*10^-7 M

this is the concentration of Br-

then

Ksp = Ag*br

[Ag+] = Ksp/Br = (5.35*10^-13)/(6.27*10^-7) = 8.53269537*10^-7

c)

%remaining = (8.53269537*10^-7)/(0.01) * 100 = 0.008532 %

NOTE: consider that Ksp values vary depending on literature

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.