Answer to Question #128773 in General Chemistry for Jocabed Jimenez

Question

Determine the molar solubility of Ba(IO3)2 Kps Ba(IO3)2 = 6.0 x 10-10

A saturated solution of bismuth (III) sulfide, Bi2S3, is held in water at 25°C. Analysis of the solution at equilibrium shows that the bismuth ion concentration is 3.4 x 10-15 M. Assuming that Bi2S3 is completely dissociated in water, calculate Kps for this compound.



Sulfuric acid is a diprotic acid. Predict the pH that a solution of potassium hydrogen sulphite, KHSO3, will have when dissolved in water. (Ka1 = 1.7 x 10-2; Ka2 = 6.4 x10-8)

Answer 1

1 .

Let  molar solubility of Ba(IO3)₂ is " S"

KS_p = [Ba²⁺] . [ IO₃^-]² .

6 * 10^-10 = 4S³.

So , S = ( 0.15 * 10^-9 )^{1/3 .}

So , S = 0.55 * 10^-3 . = 5.5 * 10^-4 .

2 .

 bismuth ion concentration is 3.4 x 10-15 M.

A saturated solution of bismuth (III) sulfide, Bi₂S₃

Let Solubility of Bi₂S₃ is "S"

THEN 2S = 3.4 x 10^-15

S= 1.7 x 10 ^-15

[Bi³⁺] = 2S .

[S^2-] = 3S .

Ksp = [ Bi³⁺]² . [S^2- ]³.

= ( 3.4 X 10^-15 )² . ( 5.1 X 10^-17 )³ .

= 1533.44 X 10^-66 . = 15.33 X 10 ^-64 .

3 .

P^H = p^K1 + P^K2 . / 2 .

p^H = ( 2 - log1.7 ) + ( 8 - log6.4 ) / 2 .

p^H = 5 - log1.7 / 2 - log6.4 / 2 .

pH = 5 - 0.23044 / 2 - .80617 / 2 .

p^H = 5 - 0.115 - 0.4030 = 4.482