Question

Question 20 Silver chromate, Ag Cros, has a Ksp of 8.96 x 10-12 Calculate the molar solubility in mol/L of silver chromate. 5.61 x 105 1.31 x 104M 1.65 x 10'M 2.29 x 104 2.25 x 10-12M

Answer 1

Answer- Let's start with the dissociation reaction of silver chromate:

Ag₂CrO₄ (s) $\rightleftharpoons$ 2Ag⁺ (aq) + CrO^2-₄ (aq)

Consider 'x' to be the molar solubility of the ions present in silver chromate.

So, Ksp can be written as,

Ksp = [2 Ag⁺]² [ CrO^2-₄ ]

Ksp = (2x)². x

Ksp = 4x³

Ksp = 8.96 x 10^-12

x³ = 8.96 x 10^-12 / 4

x³ = 2.24 x 10^-12

On solving, we get = 1.30843 x 10^-4 M $\simeq$ 1.31 x 10^-4 M

Thus, the correct option is 1.31 x 10^-4M

Answer 2

Naw it’s like atleast 2 or something