1)A solution contains 8.65×10-3 M silver acetate and 9.83×10-3 M chromium(III) nitrate. Solid sodium hydroxide is...

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?

formula =

B. What is the concentration of hydroxide ion when this precipitation first begins?
[OH^-] = _____M

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?

formula =

_____

B. What is the concentration of bromide ion when this precipitation first begins?
[Br^-] = _____M

Solutions

Expert Solution

1A) The possible compounds that can precipitate are silver oxide, Ag₂O (K_sp = 1.52*10^-8) and chromium hydroxide, Cr(OH)₃ (K_sp = 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) + O^2- (aq)

K_sp = [Ag⁺]²[O^2-] = (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)

K_sp = [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).

queen_honey_blossom answered 1 year ago

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