The value of K sp for Cd(OH) 2 is 2.5 × 10 –14 . (b) The solubility of Cd(OH) 2 can be increased through formation of the complex ion CdBr 4 2- (K f = 5 × 10 3 ). If solid Cd(OH) 2 is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of Cd(OH) 2 to 1.0 × 10 -3 mol/L?

Hello everyone. So in this video we want to go ahead and determine the initial amount of K C L. So we're gonna go ahead and first right out the equations of our molecules and related to its K. S. P and K. F values given to us in the problem. So first one is gonna be of our h g O H two. So that's going to be our starting material. And of course we have our equilibrium arrows, it's going to go and associate into our H G two plus Canadians as well as two moles of hydroxide and ions. And that K. S. P value is going to be 3.1 times 10 to the negative 26. As for the formation of our complex ion, that's going to be when H G two plus reacts with four moles of cl minus and ions. Again we have our equilibrium arrows is gonna go ahead and create that complex ion of h g c l 42 minus. And the constant for that the K F value is 1. times 10 to the 16 power. So we're basically gonna go ahead and add these two reactions up. We can first see that the HG two plus will go ahead and cancel and whatever remains will just go ahead and add that up. So for our left side of the equation we have HG O H two which is a solid acting with four moles of cl minus and then we're gonna go ahead and add the broom arrows. And on the right side of equation we have the complex ion H g c l 42 minus. And these two moles of hydroxide and now we still need a constant. And that constant is when our ksB is multiplied by r k F. Giving us the value of 3. times 10 to the negative 10. We can go ahead and assume that all, that's all you all the soluble h g two plus from the h g O h two is converted to h g c L four. So this complex ion here, so then we can go ahead and say the concentration of h g c l 42 minus is equal to 1. times 10 to the negative three. And so our concentration of o h minus is equal to two times 1.12 times 10 to the negative three. Putting that into my calculator, I get the value of 2.4 times 10 to the negative three from a concentration of hydroxide. So I'm gonna go ahead and actually scroll down right here to give us a little bit more space. So we're gonna go ahead and let c equal to the initial concentration of r k c L, which is equal to the initial concentration of our chloride and ions. So at equilibrium I'll just put that E or E Q. Just to make it shorter, the concentration of our cl minus is going to equal to c -4 times 1.2 times 10 to the -3. If we go ahead and simplify the right side right here we get that this equals to C -4.8 times 10 to the -3. So again we're gonna go ahead and bring up the equation for R. K value which is we said to be 3.41 times 10 to the negative 10. Again that's going to be the concentration of our products over our reactive and not including any liquids or new solids. So on top we're gonna go ahead and then have hg cl 42 minus and O. H minus and the constant in front of our O. H. And the equation is now going to be the power. So that's of two is going to be all over the concentration over cl minus to the fourth power. So now I'm gonna go ahead and actually insert numerical value, so I can actually get some sort of value. So for my concentration of the complex ion I have 1.2 times 10 to the negative three. And for my concentration of my hydroxide I'll have 2. times 10 to the negative three squared. And of course we have the concentration of our chlorine and ions which is c minus 4.8 times 10 to the negative three to the power of four. And now we can actually assume that the sea or the initial concentration is going to be large relative to this 4.8 times 10 to the negative three value. And so we can basically just ignore this value then because this is very big and this amount of um this number here will not affect this concentration. So we're gonna go ahead and simplify that. So 3.41 times 10 to the negative 10 were equal to 1.2 times 10 to the negative three, multiplied by 2.4 times 10 to the negative three squared. And because we're ignoring that um 4.8 times 10 negative three value, we'll just have C. To the power of four. So now we're basically going to go ahead and solve for c. four or specific. Just see. And if you do these rearrangements and mathematical manipulation, we get that the value of R. C. Is equal 2.1-18 moller. And then we go ahead and around this up to its appropriate amount of significant figures, we get 2.1 molar. So then my final answer is that the initial amount of K. C. L. Needed to increase the molars reliability of the solution is equal to 2.1. So initial amount of K. C. L. Or equal to 2.1 molar. And this right here is going to be my final answer for this problem. Thank you all so much for watching

Ch.17 - Additional Aspects of Aqueous Equilibria

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Brown 15th Edition

Ch.17 - Additional Aspects of Aqueous Equilibria

Problem 109b

Chapter 17, Problem 109b

The value of K_sp for Cd(OH)₂ is 2.5 × 10^–14. (b) The solubility of Cd(OH)₂ can be increased through formation of the complex ion CdBr₄^2- (K_f = 5 × 10³). If solid Cd(OH)₂ is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of Cd(OH)₂ to 1.0 × 10^-3 mol/L?

Verified step by step guidance

Identify the dissolution reaction for Cd(OH)2: \[ \text{Cd(OH)}_2 (s) \rightleftharpoons \text{Cd}^{2+} (aq) + 2\text{OH}^- (aq) \]

Write the expression for the solubility product (Ksp) of Cd(OH)2: \[ K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2 \]

Consider the formation of the complex ion \( \text{CdBr}_4^{2-} \) and write the formation reaction: \[ \text{Cd}^{2+} + 4\text{Br}^- \rightleftharpoons \text{CdBr}_4^{2-} \]

Write the expression for the formation constant (Kf) of the complex ion: \[ K_f = \frac{[\text{CdBr}_4^{2-}]}{[\text{Cd}^{2+}][\text{Br}^-]^4} \]

Set up the equilibrium expressions and solve for the initial concentration of NaBr needed to achieve the desired solubility of Cd(OH)2, considering the common ion effect and the complex formation.

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Key Concepts

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

Solubility Product Constant (Ksp)

Ksp is an equilibrium constant that quantifies the solubility of a sparingly soluble ionic compound. It is defined as the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced equation. For Cd(OH)2, Ksp = [Cd^2+][OH^-]^2, and a lower Ksp indicates lower solubility. Understanding Ksp is essential for predicting how changes in concentration affect solubility.

Complex Ion Formation

Complex ion formation occurs when a metal ion binds with one or more ligands, resulting in a more soluble species. In this case, the formation of the complex ion CdBr4^2- significantly increases the solubility of Cd(OH)2 in solution. The stability constant (Kf) quantifies the strength of this complex, and a high Kf value indicates that the complex is favored in solution, enhancing the solubility of the original compound.

Le Chatelier's Principle

Le Chatelier's Principle states that if a system at equilibrium is disturbed, the system will adjust to counteract the disturbance and restore a new equilibrium. In the context of solubility, adding NaBr increases the concentration of Br- ions, which can shift the equilibrium of the dissolution of Cd(OH)2, thereby increasing its solubility. This principle is crucial for understanding how changes in concentration affect the solubility of ionic compounds.