Dissolution of 5.0 x 10^-3 mol of Cr(OH)3 in 1.0 L of 1.0 M NaOH gives a solution of the complex ion [Cr(OH)4]- (Kf = 8 x10^29). What fraction of the chromium in such a solution is present as uncomplexed Cr3+?

Video transcript

Welcome back everyone when 2.50 times 10 to the negative third power moles of copper chloride are added to 450 mL of a 4500.250 moller potassium cyanide solution. A complex ion with the formula copper to cyanide. An ion is formed. The formation constant for the complex ion is one point oh times 10 to the 16th hour. What is the fraction of complex copper ions in the solution? Now make note of the fact that potassium cyanide is considered the salt of a strong base and weak acid. The strong base we would recognize as potassium hydroxide and the weak acid we would recognize as H. C. N. Or hydrogen cyanide. Now because we have our strong base we can go ahead and consider potassium cyanide as our solvent and this strong base or this solvent rather is going to dissociate fully so that we form K plus plus C. N minus. And so now we need to focus on our solid which is our copper chloride. Recognize that this is considered a weak electrolyte. And so this is only going to partially disassociate. And so rather than a regular arrow, we should actually have a equilibrium arrow where we would form our copper plus one catalon and our chloride -1 and ion. Now in order to get to our fraction of uncomplicated copper cat ions, we need to figure out the polarity or molar concentration of are added copper cat ions. And so we would find that by taking the given moller concentration or rather the given number of moles of our copper one chloride solid here. So that's given as 2.50 times 10 to the negative third power most of copper one chloride divided by our volume of our solution. Which we are given as sorry not the volume but the Volume here is given in ml. And so we need this to be in terms of leaders. So we're given 450 ml converting to leaders. We have million. The denominator leaders in the numerator are prefix milli we recall tells us that we have 10 to the third. Middle leaders equivalent to one leader canceling all male leaders were left with a volume now in leaders. And so we have moles per liter as our final units which we recall is interpreted as a polarity. And so We result with a molar concentration equal to 0. Mohler. Now we can move on to show our formation equation for our complex. And so we have our copper cat ion which reacts with our cyanide, C N minus an ion. And in an equilibrium we form the complex copper to cyanide. So C. N in parentheses subscript of two and this is a plus or minus one an ion Balancing this out, we need to place a coefficient of two in front of our cyanide. We can interpret our concentrations in polarity as our initial concentrations. And so we would say that we have initially for copper 0. Present initially. And then for cyanide we can take it smaller concentration from the prompt as 0.250 initially. Now for our complex we have zero of that formed initially because it's a product. Now for the change we're going to assume that we have 100% completion of the reaction. And so the change would be minus are smaller amounts of our reactant which is our copper carry on. It's limiting. And so we would have minus and sorry, we would have -0. from both of our reactant. Now for our second reactant cyanide we have a coefficient of two. So we're going to actually have negative two Times the change there for our product, we would have plus that change since it's forming. So plus 0.005556. Again, this is under the assumption that 100% of our reaction occurs in the forward direction. Now we would have a small chance of a back reaction which we can account for here. Where or we can say the reverse reaction rather where the change here would be the opposite. So we would have plus our copper caddy on and then we would have plus two X. Of our cyanide and then minus X. For our complex. Now focusing on what occurs at equilibrium. And actually we skipped a row here. So before we write out the reverse reaction potential, we need to consider what happens when the reaction proceeds in the forward direction at 100%. So this would be after the reaction completes in the four direction, we would have at no longer equilibrium. Actually will say f here for final Concentrations we would have zero left of our copper caddy on. We would have 0. Rather, Let's say 238889. And then for our complex we would have 0.005556. So now going back to our equilibrium, once we have the reverse reaction we would just carry our variables down. So we would have at equilibrium X. Of our copper carry on. 0.23 Sorry 0. 8889 plus two X. For our cyanide. And then 0. minus X. For our complex. And so now writing out the equilibrium expression for our formation constant K. F. We want to take products over reactant. So for our product we have the concentration of our complex. See you CN sorry, this is C. U. Parentheses C. N. Sub two. And then we have a plus one counting on charge. This is our only product divided by our reactant for our reactant, we have copper caddy on which is a plus one catty on concentration multiplied by the concentration of cyanide which again has a coefficient of two. So we raise it to a power of two, plugging in what we have for our equilibrium. And plugging in the given formation constant for the complex ion were given in the prompt, the value 1.0 times 10 to the 16th power for K. F. Which is equal to our numerator for the concentration of our complex at equilibrium. We determined to be 0.55556. Or sorry, it should have 3/5. So let's change that here. This is minus X. In our denominator we have our concentration of copper caddy on at equilibrium as just X. And in our our second term in our denominator which is our concentration of cyanide at equilibrium is 0.238889 plus two X. And this is squared. Let's make note of the fact that we have a very high value of K. F. And so therefore X. Is negligible. And so we can simplify this expression so that we have now one point oh times 10 to the 16 power, our formation constant given in the prompt set equal to 0. minus. Or sorry, just 556. Because we are ruling out the ex Now since it's negligible, this is divided by X. Which is multiplied by 0. which is squared. So we've rolled out the X terms with the algebraic signs in front of them. We've rolled out this X. And we've rolled out the plus two X. Here. Now with this simplification we can solve for our variable X. To determine our concentration of copper caddy on. So we have X. Which is going to equal our numerator 0.005556. Now divided by our formation constant 1.00 times 10 to the 16th power multiplied by . which is squared. Now solving for X. We would find that X is equal to 9.73576 times 10 to the negative 18th power. And now we need to figure out our fraction of uncomplicated copper cat ions. And so we would take our concentration of copper caddy on at equilibrium and we're dividing that by our initial concentration of copper cat ion. And so we would refer to our value for the denominator from our initial step, which was calculating the polarity of our added copper Catalans. We determined that above as 0. Mohler, sorry 5556 Mohler here. So let's write that neatly and then our numerator, we just solved as 9.73576 times 10 to the negative 18th power. So simplifying this quotient in our calculators. We would come up with a fraction of our uncomplicated copper ions equal to 1.75 to 3 times 10 to the negative 15th power. Now our numerator here for the concentration of copper Catalan and equilibrium is in units of polarity to and our polarity units cancel out. And so this fraction here is our final answer again as the fraction of our complex copper Catalans. So this corresponds to choice D in the multiple choice as our final answer. I hope everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.