What is the solubility of CN − ions in a solution of 5.5 M Hg 2 (CN) 2 , with a K sp of 5.0 × 10 − 40 ?

Here we're asked, what is the solubility of cyanide ions in a solution that's 5.5 molar for mercury 1 cyanide? Here it has a Ksp value of 5.0 times 10 to the minus 40. Alright. So since we're talking about Ksp, we're dealing with an ionic solid, and we have to break it up into its ions. We're going to say here that this is mercury 1, which is h g h g 22+aqueousplus2cynonines. Now, this is mercury 1 because it's 2 mercury ones combined together, so their collective charge together is 2 plus. Okay? We say that this is called a dimer, so it's basically 2 similar structures, bonded together. No, you don't need to know this for the exam or for homework, it's just a little bit of extra information. Now, here we're going to say we have initial change equilibrium. The molarity there is cool, but we don't really need to use it because we're looking for the solubility of the cyanide ion. Alright. So here we're going to say, in an ice chart we ignore solids and liquids, so the solid reactant is ignored. Initially, we have 0 and 0. Now, we're going to say that we have to lose reactant to make product. So plus x, and this is a 2, plus 2 x. Plus x, plus 2 x. Alright. So now we're gonna say KSP equals products over reactants, but the reactant is a solid, so we ignore it. So Ksp just equals products. So it equals mercury 1 times cn minus squared. Ksp here is 5.0 times 10 to the minus 40 equals x times 2x squared. So this is x times 4x squared, so this equals 4x cubed. And along each path, each step, it's still equal to k s p. So what we're gonna do now is we're going to divide both sides by 4, and we're gonna bring the math over here. So now we're gonna say 1.25 times 10 to the minus 40 equals x cubed. Take the cube root of both sides here and that'll give me my x. So x here equals 5.0 times 10 to the minus 14. But we found x, that represents the solubility of the ionic solid. They're asking us for an ion. So look at the equilibrium row, and they want us to find it for cyanide ions. At equilibrium, cyanide ion equals 2 x. So you would say that the concentration of cyanide ion equals 2x. We just found out what x is, so plug it in. So this equals 1.0 times 10 to the minus 13 molar. So remember, solubility is referencing molarity. So this would be the concentration or molar solubility of the cyanide ion within this given question.

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18. Aqueous Equilibrium

Solubility Product Constant: Ksp

General Chemistry

18. Aqueous Equilibrium

Solubility Product Constant: Ksp

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Multiple Choice

What is the solubility of CN⁻ ions in a solution of 5.5 M Hg₂(CN)₂, with a K_sp of 5.0 × 10⁻⁴⁰?

4.5× 10⁻²⁰ M

5.0 × 10⁻¹⁴ M

1.1 × 10⁻²⁰ M

1.0 × 10⁻¹³ M

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Verified step by step guidance

Understand the dissociation of Hg2(CN)2 in water. The compound dissociates into Hg2^2+ and CN^- ions. The balanced equation is: Hg2(CN)2 ⇌ Hg2^2+ + 2CN^-.

Identify the expression for the solubility product constant (Ksp). For the dissociation of Hg2(CN)2, the Ksp expression is: Ksp = [Hg2^2+][CN^-]^2.

Let 's' be the solubility of Hg2(CN)2 in mol/L. Since each formula unit of Hg2(CN)2 produces one Hg2^2+ ion and two CN^- ions, [Hg2^2+] = s and [CN^-] = 2s.

Substitute the concentrations into the Ksp expression: Ksp = (s)(2s)^2 = 4s^3.

Solve for 's' using the given Ksp value: 5.0 × 10^-40 = 4s^3. Rearrange to find s, which represents the solubility of CN^- ions.