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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

Respuesta :

sebassandin

Answer:

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

Explanation:

Hello,

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

[tex]Ag_2CrO_4(S)\rightleftharpoons 2Ag^+(aq)+CrO_4^{-2}(aq)[/tex]

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

[tex]Ksp=[Ag]^2[CrO_4^{-2}][/tex]

Thus, given the change [tex]x[/tex] due to the dissolution of silver chromate, we obtain:

[tex]1.2x10^{-12}=(2x)^2\times x[/tex]

Hence, we solve for [tex]x[/tex]:

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

Thus, the concentration of silver ions will be:

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

Best regards.

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