What is the molar solubility of BaF2 in a solution containing 0.0750 M LiF (Ksp = 1.7 x 10^-6)
(a) 2.3 x 10^-5 M
(b) 3.0 x 10^-4 M
(c) 1.2 x 10^-2 M
(d) 1.3 x 10^-3 M

Question

Accepted Answer

Welcome back everyone. We need to calculate the molar scalability of lead to bromide with a solid ability product constant equal to 6.6 times 10 to the negative six power and 0.1 50 moller Aquarius potassium bromide solution. So let's begin with our solid lead to bromide. And sorry P B B R sub two. It's a solid and so we're going to undergo a dissolution to form its ions in which are solid is an equilibrium with its ions. Where we have the lead two plus catalon form as well as R B R minus an eye on form. We need to make sure that this is balanced. And so we'll have to place a coefficient of two in front of our bromide. An ion product because we have two moles of bromide on the reactant side. Now that we have our balanced dissolution reaction, we need to make an ice chart to help us calculate our molars liability. And so we have I C. E. For our I start where for our initial amounts and sorry by the way we do not include our products. Sorry we do not include our solid re agents. We only include our acquis three agents. So our ions. So for the initial amount of our ion products for our lead carry on we would put zero and that's because we only know our initial concentration of our bromide an ion given that we are told the concentration of potassium bromide As 0.150 moller. We know that this is true because we have potassium bromide a part of our solution. And so we would have the presence of potassium plus one cat ions and B r minus an ions. And if we have B r minus an ion in solution, that means we can assume that for our lead to bromide we would have the same initial concentration. So we would plug that in as 10.1 50 dolar. Now for our change of our products we would have plus both amounts. So in this case we have a coefficient of one in front of lead two plus. So we have plus X. And then for our two moles of bromide an ion we have plus two X. As our change now going to our equilibrium we would carry everything down. So we have X. For lead two plus and for bromide we have 0.150 and plus two X. Now let's consider whether our plus two X. Is going to be negligible, recall that we can determine if X is negligible by taking the concentration of our bromide, an ion divided by our K. S. P. Of our solid. And so we have 0.150 moller divided by R given K. S. P. Of 6.6 times 10 to the negative six power. This will result in a value of 22,727 which is significantly greater than our comparison being 500. And so we would say that there four plus two X. It's definitely negligible. And so we can rule this out from our equilibrium, meaning that we know our equilibrium concentration of our bromide and ion. Recall that we can calculate R. K. S. P. By taking the product of our ion concentrations and solution. And so for our ion products we have our lead two plus Catalan which we have just one mole of. And then we have for our bromide an ion a coefficient of two. So we would have br minus raised to a power of two because recall that in our K. Sp equation, the coefficients from our reaction become exponents. So maybe to make it clear let's make this exponents pink. Now let's plug in what we know and solve for our moller Celje bility. So plugging in our given K. S. P. For lead to bromide. We have 6.6 times 10 to the negative six power as given in the prompt set equal to our concentration of lead Catalan which is considered X at equilibrium. So we're solving for X multiplied by our concentration of bromide which we just determined to be 60.150 Which is raised to a power of two. So continuing to simplify this, we would have the same left hand side Of our equation set equal to X. multiplied by .150 raised to a square power which results in a value of 0.0225. And so simplifying for X. We would Divide both sides by 0.0225. So that it cancels out on the right hand side. And we will find that our value X. Is equal to a value of 2.9 times 10 to the negative fourth power moller. And notice that we only use 266 here because our minimum number of sick fix given an R K. S. P figure is two. And so we've successfully determined our solar cell ability which was or which is the maximum concentration of our lead two plus cad ion in solution before it begins to precipitate out of solution. So this answer X equals 2.9 times 10 to the negative fourth power moller will correspond to choice C. As our final answer. To complete this example for the molars eligibility of lead to bromide in 100.1 50 moller. Aquarius potassium bromide solution. 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. And sorry, C should be highlighted here. Okay, see everyone in the next video

What is the molar solubility of BaF2 in a solution containing 0.0750 M LiF (Ksp = 1.7 x 10^-6) (a) 2.3 x 10^-5 M (b) 3.0 x 10^-4 M (c) 1.2 x 10^-2 M (d) 1.3 x 10^-3 M

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Key Concepts

Molar Solubility

Solubility Product Constant (Ksp)

Common Ion Effect