Question

What is the concentration of the silver ion in silver chromate, Ag₂CrO₄, if its solubility product constant (Kₛₚ) is 1.2 x 10⁻¹². Hint: write the equation first! *
2 points

1.4 x 10⁻⁵

1.1 x 10⁻⁹

1.6 x 10⁻¹²

2.4 x 10⁻¹²

Answer 1

Answer:

$[Ag^+]=1.3x10^{-4}M$

Explanation:

Hello,

In this case, given the solubility product of silver chromate:

$Ag_2CrO_4(S)\rightleftharpoons 2Ag^+(aq)+CrO_4^{-2}(aq)$

Now, the law of mass action excluding silver chromate as it is solid, turns out:

$Ksp=[Ag]^2[CrO_4^{-2}]$

Thus, given the change $x$ due to the dissolution of silver chromate, we obtain:

$1.2x10^{-12}=(2x)^2\times x$

Hence, we solve for $x$:

$x=\sqrt[3]{\frac{1.2x10^{-12}}{2^2} } = 6.7x10^{-5}M$

Thus, the concentration of silver ions will be:

$[Ag^+]=2x=2*6.7x10^{-5}M=1.3x10^{-4}M$

Best regards.