Is the solubility of Zn(OH)2 increased, decreased, or unchanged on addition of each of the following substances? Write a balanced net ionic equation for each dissolution reaction. (See Appendix C.6 for formulas of complex ions.) (c) NaCN

Question

Accepted Answer

Welcome back everyone. What will happen to the salt ability of iron three hydroxide when potassium cyanide is added to the solution provide the balanced net ionic equation for the dissolution process. And our hint is to refer to complex ions formations. So let's recall that when we have potassium cyanide added to our solution With Iron three hydroxide, we will have the formation of the complex being iron six cyanide with a three minus charge. So this is an an ion complex and we are given a formation constant for this complex equal to four times 10 to the 43rd power. Meaning that this complex is very likely to form when these compounds are in our solution together. This also means that we have an increased Celje bility of our salt iron three hydroxide. So recall that this is going to undergo dissolution where we will form an equilibrium between iron three hydroxide and its ion products forming the iron three plus cat ion and our hydroxide an ion, we need to make sure this is balanced, that will place a coefficient of three in front of hydroxide on the product side. And now focusing on our potassium cyanide, which is our other salt which will dissolve into the potassium plus one carry on. And our cyanide one minus an ion. Now note that based on our complex ion formation, which we said has a formation constant of four times 10 to the 43rd power indicating that our complex iron six cyanide three minus will definitely form. This means that the ions present in this complex are going to react. And so we can write out a 30 equation where we have our iron three plus Catalan reacting with our cyanide, an ion in equilibrium to form our complex iron six cyanide three minus. Now, we need to make sure things are balanced. Note that in our complex, we have six moles of cyanide. And so we're going to need to take the second equation. We wrote for the dissolution of potassium cyanide and multiply everything by six. And for our third equation, we need to make sure that we have a coefficient of six in front of our cyanide an ion because we need six moles to match with the six moles of our cyanide on the product side. Let's also make sure we have everything labeled. So our salt should be labeled with solid. And now we're going to add up each of these equations where we can realize that we can cancel out our iron carry on from our third equation with the iron caddy on, on our product side of the first equation and we can also cancel out the six moles of cyanide when we multiply the second equation by six with the six moles of cyanide on the react inside of our 30 equation. Now, we're just going to carry everything down that we have left to come up with our overall equation as our net ionic equation. So what we would have is iron three hydroxide, Our salt which is added to six moles of potassium cyanide. And they react in equilibrium to form our products where we have our complex iron, six cyanide, three minus an ion, Six moles of potassium cat ions and three moles of hydroxide. And actually, I just remembered that this is not going to be our net ionic equation just yet because we should recall that our acquis species are going to disassociate. And so out of the acquis species that we have, which can disassociate, we would rewrite our equation so that we have our iron three hydroxide plus six moles of potassium cat ions plus six moles of cyanide and ions in equilibrium with our complex product iron six cyanide three minus an eye on Plus our six moles of potassium Canadians plus three moles of hydroxide. And so, canceling out our common ion, we have six moles of potassium carry ons on both sides of the equation. And so now we can finally confirm that we have our net ionic equation In which we have our salt iron three hydroxide Plus six moles of cyanide in equilibrium with the complex iron six cyanide with a three minus an ion charge plus our three moles of hydroxide. And they should have a -1 charge here. So overall, we have determined our second answer as our balanced net ionic equation here highlighted in yellow. And we've also determined our first answer above in which based on the reaction between potassium cyanide and iron three hydroxide for the formation of our complex with its formation constant of four times 10 to the 43rd power based on this very high formation constant, we will confirm that are solid ability of our salt iron three hydroxide increases. And so the formation of our complex ultimately increases the salt ability of our salt iron three hydroxide. So what we have highlighted in yellow are our two final answers to complete this example corresponding to choice B in the multiple choice. I hope everything I reviewed was clear. If you have any questions, please leave them down below. And I'll see everyone in the next practice video.

Is the solubility of Zn(OH)2 increased, decreased, or unchanged on addition of each of the following substances? Write a balanced net ionic equation for each dissolution reaction. (See Appendix C.6 for formulas of complex ions.) (c) NaCN

Verified Solution

Key Concepts

Solubility Product Constant (Ksp)

Common Ion Effect

Complex Ion Formation