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A solution is 0.010$M$ in both $\mathrm{Cu}^{2+}$ and $\mathrm{Cd}^{2+}$ . What percentage of $\mathrm{Cd}^{2+}$ remains in the solution when 99.9$\%$ of the $\mathrm{Cu}^{2+}$ has been precipitated as CuS by adding sulfide?

$4.0 \times 10^{1} \mathrm{M}=40 \mathrm{M}$

Chemistry 102

Chapter 15

Equilibria of Other Reaction Classes

Chemical Equilibrium

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Hello, everyone. This is reeking And today working on problem 43. So we're told the solution is went 010 Moeller. Huh? Coffer, Count me in and we want to know the percentage Cadman after 99.9% of copper has been precipitated Wave so far two. All right, So if 99.9% of copper too has been precipitated, that only 0.1% 1 is a solution. So we can calculate concentration of copper as point one over 100 just the amount that remains Times 0.1 Bulls for leader which is the concentration originally get one times 10 to the negative Fifth Moeller So see you us as a K s P of 6.7% rate of 42 cadmium sulphide as a K A s p of 2.8 times 10 to the negative 35 given the following conditions. So now we want to calculate the concentration of so fight I owns for both So K S P is equal to copper. It's so fired 6.7 times and 42 equals one time. Since the negative fifth times X X is equal to 647 times 10 to 7 Mullah waas sulfide I own concentration. Now plug this and you are disassociation equation for cadmium or are cast before canyon to see the plus two. This minus two sell for a concentration canyon 2.8 times Oh, for 6.7 times center the native 37th. Um, when we get 40 Moeller um and so what this means is the concentration of cadmium can increase to around 40 Moeller before it will precipitate. And the concentration of cadmium is only 0.10 Moeller. Right now 40 is way, way, way, way, way larger than point. Oh, Moeller so, uh, safe to assume that 100% of CD was too has dissolved. So I hope this video was helpful and I'll see you in the next

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