Calculate the molar solubility of Ni(OH) 2 when buffered at pH (b) 10.0.

hi everyone for this problem it reads what is the soluble itty and grams per liter of nickel to hydroxide with a solid ability product of six times 10 to the negative 16 in a solution buffered at ph equals 5.85. Okay, so our goal here is to find the sala bility in grams per leader. All right. And so let's go ahead and start off by writing our equilibrium that is established between are solid and the ions and solution. Okay, So we have our nickel to hydroxide and this is our solid. And at equilibrium it produces nickel ions and hydroxide ions. And were given the value of the equilibrium constant K. S. P. And Casspi represents the product of the molar concentrations of our ions. Okay, so this product is going to be our nickel ions times are hydroxide ions and we're paying attention to our coefficients here because we have two hydroxide, we're going to square this here And in the problem we are told what r K. S. P. Value is and that value is 6. times 10 to the negative 16. Okay. Our goal here is to calculate the sala ability and grams per leader. Okay, and so here with our K. S. P value in order for us to solve for solid ability, we're going to want to figure out what is our concentration of hydroxide. Okay, and were able to calculate our concentration of hydroxide from one key thing that's given in the problem. Okay, so if we can remember our concentration of hydroxide is equal to 10 to the negative p. O. H. Okay, so remember we have a ph scale and we have a P O. H scale. And so our goal here is to solve for the concentration of hydroxide. And one thing to remember is that P O H is equal to 14 minus P. H. All right. And we know what our ph is because we're told in the problem. So our P O. H. Is going to be 14 minus 5.85, which gives us a P. O. H. Of 8.15. So let's go ahead and plug that in. So we get 10 to the negative P O. H. We said our P O H is 8.15. Okay, so that gives us a hydroxide concentration of 7.795. Oops, put a lot of space there. So 7. times 10 to the -9 moller. So this is our hydroxide concentration. So let's go ahead and solve for our nickel concentration. Okay, so our concentration of nickel ions we're gonna say is equal to X. Okay, and from here we're going to solve for X. Okay, so we know that we have our K sp expression is six point oh times 10. So the negative 16 is equal to X. Which is our concentration of nickel times our concentration of hydroxide which we just solved for. So 7. Times 10 to the negative nine squared. Okay, so this is basically this rewritten. Alright, but we're substituting s for a concentration of nickel. Alright, so our goal here is to solve now for X. Alright. So we'll first isolate it by dividing or isolating X by dividing both sides by 7.0795 times 10 to the -9 squared. All right. And when we do that, we get 11.971 molar. So let's go ahead and write that here are concentration of nickel which we let X be its representative is equal to 11.971 molar. Alright, so now that we know our concentration of hydroxide and we know our concentration of nickel. We can calculate the solid ability. Alright. And we can calculate the solid ability because our concentration of nickel ion is equal is the same thing as our concentration of our nickel to hydroxide. Okay, so by knowing what our concentration of nickel is, we'll be able to figure out our soluble itty. So let's go ahead and we know that our concentration of nickel ions is 11.971 Moeller. Remember molar is moles over leader and it's specifically moles of nickel. Two hydroxide. Okay, so we want to go from moles per leader. two g per leader. Okay, this is our goal. So let's go ahead and do that. Alright, so and one more of nickel to hydroxide. We need the molar mass. The molar mass is 92.71 g. So making sure our units cancel here are moles cancel and we're left in grams per leader, which is what we're looking for. All right. So once we calculate this out, we're going to get a final answer of 1.1 times to the third grams per leader. Alright, This is our final answer. This is the solid ability in grams per leader. That's it for this problem. I hope this was helpful.

Ch.17 - Additional Aspects of Aqueous Equilibria

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Brown 14th Edition

Ch.17 - Additional Aspects of Aqueous Equilibria

Problem 62b

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Chapter 17, Problem 62b

Calculate the molar solubility of Ni(OH)₂ when buffered at pH (b) 10.0.

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Identify the dissolution equation for Ni(OH)_2: Ni(OH)_2(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^{-}(aq).

Write the expression for the solubility product constant (K_{sp}) for Ni(OH)_2: K_{sp} = [Ni^{2+}][OH^{-}]^2.

Determine the concentration of OH^{-} ions from the given pH: Use the relation pOH = 14 - pH to find pOH, then calculate [OH^{-}] = 10^{-pOH}.

Substitute the [OH^{-}] value into the K_{sp} expression and solve for [Ni^{2+}], which represents the molar solubility of Ni(OH)_2.

Use the known K_{sp} value for Ni(OH)_2 to calculate the molar solubility by substituting the [OH^{-}] concentration and solving for [Ni^{2+}].

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molar Solubility

Molar solubility refers to the maximum amount of a solute that can dissolve in a given volume of solvent at a specific temperature, expressed in moles per liter (mol/L). It is a crucial concept in understanding how substances interact in solution, particularly for sparingly soluble compounds like Ni(OH)2. The molar solubility can be influenced by factors such as pH, temperature, and the presence of other ions in solution.

pH and its Effect on Solubility

pH is a measure of the acidity or basicity of a solution, which can significantly affect the solubility of certain compounds. For nickel(II) hydroxide, Ni(OH)2, an increase in pH (making the solution more basic) can enhance its solubility due to the formation of soluble nickel complexes. Understanding how pH influences the dissociation of hydroxides is essential for calculating molar solubility in buffered solutions.

Equilibrium and Ksp

The solubility product constant (Ksp) is an equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds. For Ni(OH)2, the Ksp expression relates the concentrations of the ions in solution at equilibrium. By knowing the Ksp value and the pH of the solution, one can derive the molar solubility by setting up an equilibrium expression that accounts for the concentration of hydroxide ions, which are influenced by the pH.