1)A solution contains 8.65×10 -3 M silver acetate and 9.83×10 -3 M chromium(III)
ID: 593418 • Letter: 1
Question
1)A solution contains 8.65×10-3 M silver acetate and 9.83×10-3 M chromium(III) nitrate.
Solid sodium hydroxide is added slowly to this mixture.
A. What is the formula of the substance that precipitates first?
B. What is the concentration of hydroxide ion when this precipitation first begins?
[OH-] = _____M
2)
A solution contains 5.94×10-3 M lead nitrate and 1.27×10-2 M silver acetate.
Solid ammonium bromide is added slowly to this mixture.
A. What is the formula of the substance that precipitates first?
B. What is the concentration of bromide ion when this precipitation first begins?
[Br-] = _____M
Explanation / Answer
1A) The possible compounds that can precipitate are silver oxide, Ag2O (Ksp = 1.52*10-8) and chromium hydroxide, Cr(OH)3 (Ksp = 6.3*10-31).
The two salts ionize in different manners and hence, we need to calculate the solubility of the two salts.
Ag2O (s) <======> 2 Ag+ (aq) + O2- (aq)
Ksp = [Ag+]2[O2-] = (2x)2.(x)
=====> 1.52*10-8 = 4x3
=====> x3 = 3.8*10-9
=====> x = 6.16*10-5
Cr(OH)3 (s) <======> Cr3+ (aq) + 3 OH- (aq)
Ksp = [Cr3+][OH-]3 = (x).(3x)3
=====> 6.3*10-31 = 27x4
=====> x4 = 2.3333*10-32
=====> x = 1.2359*10-8 1.23*10-8
The solubilities of Ag2O and Cr(OH)3 in water are 6.16*10-5 M and 1.23*10-8 M respectively. Ag2O is more soluble in water and hence, Cr(OH)3 will precipitate out first (ans).
B) Cr(OH)3 is the first to precipitate and we need to find out the concentration of OH- when Cr(OH)3 begins to precipitate. We have,
6.3*10-31 = (9.83*10-3).(x’)3
=====> x’3 = 6.4089*10-28
=====> x’ = 8.6217*10-10 8.62*10-10
The OH- concentration when Cr(OH)3just begins to precipitate is 8.62*10-10 M (ans).
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