Calculate the molar solubility of Cr(OH)3 in 0.50 M NaOH; Kf for Cr(OH)4- is 8 x 10^29.

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Hey everyone, we're asked what is the Mueller scalability of silver cyanide in one point oh molar sodium cyanide solution and were given the following values. Which is our K. S. P. R. Scalability product constant for silver cyanide and our moral freezing point depression constant for our silver cyanide complex. Now when we look at sodium cyanide we know that this is going to dissociate into our sodium ions plus our cyanide ions. But since sodium is a neutral ion this is not involved in our reaction. So our sodium does not affect our scalability. Now let's go ahead and figure out our overall reaction. So we know that we have our silver cyanide and this is in its solid state and on our product side we're going to get our silver ion plus our cyanide ion. Next, in order to form our silver cyanide complex, we have our silver ion plus our cyanide ion. And this is going to form our silver cyanide complex. Now, in order to balance this out, we need to add a coefficient of two prior to our cyanide ion in our reactant side. Now we can go ahead and figure out our overall reaction. So as we can see right here our silver ion is going to cancel out in our 1st and 2nd reaction. Next we have our cyanide ion that's going to cancel out as well. And our overall reaction is going to come up to silver cyanide Plus one of our cyanide ion. And we end up with our silver cyanide complex. Now let's go ahead and create our ice chart. So our silver cyanide is a solid. So this will not be involved in our ice chart initially are cyanide ion had one point oh oh moller which we saw in our questions them and we had zero of our silver cyanide complex are changes of minus X on our react inside and a plus X on our product side. Since we're losing reactions and gaining products at equilibrium on our reactant side, we have one point oh oh minus X. And an X in our product side. So this means our K equilibrium is equal to X over 1.0 minus X. So how do we solve for K equilibrium? Well, we know that our K equilibrium is equal to our K. S p times our K. F. So plugging in these values, we have 6.0 times 10 to the - Times 3.0 times 10 to the 20. So this means our K equilibrium is equal to 1.8 times 10 to the four plugging this in we have 1.8 times 10 to the four is equal to x over one point oh oh minus x. Isolating our X. We have 1.8 times 10 to the four times 1. minus X equals X, solving for X. We get 18,000 - x Equals X. Putting our two exes together, we get 18,000 equals X plus 18, X. This gets us to 18,000 equals 18,001 x. Solving for X, we get 1.0 Moller which is going to be our final answer. Now, I hope this made sense and let us know if you have any questions.

Calculate the molar solubility of Cr(OH)3 in 0.50 M NaOH; Kf for Cr(OH)4- is 8 x 10^29.

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Key Concepts

Molar Solubility

Common Ion Effect

Formation Constant (Kf)