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Perform the following calculations involving concentrations of iodate ions:(a) The iodate ion concentration of a saturated solution of $\mathrm{La}\left(\mathrm{IO}_{3}\right)_{3}$ was found to be $3.1 \times 10^{-3} \mathrm{mol} / \mathrm{L}$ . Find the $K_{\mathrm{sp}}$(b) Find the concentration of iodate ions in a saturated solution of $\mathrm{Cu}\left(\mathrm{IO}_{3}\right)_{2}\left(K_{\mathrm{sp}}=7.4 \times 10^{-8}\right)$
a. $3.1 \times 10^{-11}$b. $5.3 \times 10^{-3} \mathrm{M}$
Chemistry 102
Chapter 15
Equilibria of Other Reaction Classes
Chemical Equilibrium
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Hello, everyone, this is Ricky. And today we're working on problem 53. So we have to calculate KSB for this association. Were given the concentration of I had I io date. I own, um because of the molar ratios, there's gonna be three times as much I a date as, um l a. So we can just multiply this by 1/3 to get are all a concentration, which is 1.3 times 10 to the third most handsome. All right. Now we can just plug these values into our K S P expression and solved for chaos p which is equal to flee 0.1 time stone toupee. Thank you. Right. And there's a second part to this question. Oh, find the concentration of Ida ions in a Saturn solution of super hot date CEO copper. I have me copper too. I did dissolves to see you plus two plus three minus a SP is. He worked seven four times since then. They ate right. Our expression K s P is equal to see you. Plus two times two I oh three minus squared. Because of the moment ratios, there should be twice much io the re minus. And although we don't always include this here, we should really included solved. So let's rewrite everything. Okay, that 7.4 times 10 did eighth physical T X times two X squared Well two for ex Cute X is equal to 2.64 comes to her, um, which is equal to the concentration of copper, too. There's twice as much I a day just copper twosome. Multiply this by two times two is yeah, 5.3 times 10 to the negative third. Well, er is equal to the hyo date high concentration.
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