Problem

Assuming that no equilibria other than dissolutio…

Problem 16 Medium Difficulty

Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.
(a) $\operatorname{TICl}(s)$ in 1.250$M \mathrm{HCl}$
(b) $\operatorname{PbI}_{2}(s)$ in 0.0355$M \mathrm{CaI}_{2}$
(c) $\mathrm{Ag}_{2} \mathrm{CrO}_{4}(s)$ in 0.225 $\mathrm{L}$ of a solution containing 0.856 $\mathrm{g}$ of $\mathrm{K}_{2} \mathrm{CrO}_{4}$
(d) $\mathrm{Cd}(\mathrm{OH})_{2}(s)$ in a solution buffered at a pH of 10.995

Answer

a. the percentage change in concentration of common ion is below $5 \% .$ So, the $S$ can be neglected.
b. the percentage change in concentration of common ion is below $5 \% .$ So, the $S$ can be neglected.
c. the percentage change in concentration of common ion is below $5 \% .$ So, the $S$ can be neglected.
d. the percentage change in concentration of common ion is below $5 \% .$ So, the $S$ can be neglected.

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Chemistry 102

Chemistry

Chapter 15

Equilibria of Other Reaction Classes

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Video Transcript

It's so for these problems. We want to set up an icebox because we need to determine the equilibrium concentration off the dissolved ions. So initially, oftentimes with ice boxes, our initial is zero. But in this case, we have some chloride on already in its 1.25 Moeller. So our quarry line, it's not gonna be zero. It's gonna be 1.25 are changes gonna be ex for both Valium and chloride? So that means our equilibrium concentrations would be ex for thallium and 1.25 plus X for chloride. So now we can use the equilibrium values in our K S P expression. So ke SP is just gonna be a concentration of Valium, times concentration of chloride. And we can make the assumption that K S P is so small compared to X that we can ignore X in this expression here when we're adding X to something that isn't zero, because that would give us a quadratic equation. If we plugged it into K, S P would end up with an X where term an ex term and then a constant. So to avoid that, we can make that approximation and plug in our values into the expression with that approximation. So 1.7 times 10 to the minus four is K S. P. Um, if we do this math we end up with this shouldn't be X squared. That shitty ex do the math. We end up with X equals 1.4 times 10 to the minus four, and that X is gonna be our concentration of Italian and then for chloride. The reason we can ignore the change in the initial concentration of chloride is that it would be less than 0.5% change. So if you did add X toe 1.25 which is up here, you get 1.250 plus 0.14 That changes so salty 1.251 And that's much. That percent changes much less than 0.5%. So you can just use the initial concentration given in the problem and that will apply for all of these. In fact, some of them you wouldn't even notice because of the number of significant figures. You wouldn't even notice the, um, change at all. So if you did the addition, he would still end up with the original concentration. So for lead to I and I very similar. We have an initial iodide concentration. We add, um, we had our change. It's gonna be ex for lead. It's gonna be two x for iodide because of the two in the equation that gives us thes equilibrium concentrations. A quick note that I died up here. I thought I was gonna be two times the concentration of lead to I died. That's given because there's two iodide ions for every lead to iodide unit. Then we use our K S P expression. We assume that Ke SP is much less than X. So we make that approximation. Ke sp is 1.4 times 10 to the minus eight. That's gonna give us extra 0.71 When we solve for X, we get 1.97 times 10 to the minus seven. That's the concentration of lead. And then for iodide, the percent change again would be less than 5%. We don't even notice a difference when we add this to ex terms. So it's just 0.71 for potassium chromite. The key here is that we're given Graham's and were given a volume. So we have to get to my Olara tea so we can use the molar mass here, divide by the number of leaders and that is gonna give us the concentration that we need. Now, there's a once one ratio of chrome 82 potassium crow bait. So that means our concentration is gonna be valid for the chromite ion plugging into the expression. Um, we'll get into the icebox rather two x for silver X for chromite. We apply our approximation and plug into the K S P expression. This time we do have a squared here. So, um, we have to first divide by point 0196 and then divide by four and then take the square root. So that gives us our X value. Now this for silver. Remember, it's two x, so we have to multiply it by two. And that gives us 2.15 times heads and minus five Moeller pro mate again, If we do the addition of X, we don't even notice the change. It also it's gonna be, um, actually, this is written incorrectly. It's gonna be 0.196 Moeller saying was the initial concentration given in the problem for cadmium hydroxide. This time we have a pH up 10.995 So that means there's gonna be, ah, fairly high concentration of hydroxide so we can get P O. H. From Ph. That's 14 minus the pH. So P O. H is gonna be 3.5 We then take Ted into the minus P o. H. That gives us our hydroxide concentration. 9.89 tends to into the minus four. That's our initial concentration of hydroxide. So that goes here in the initial two X for hydroxide extra cadmium plug into the K S P expression and ignore this two ex term. So that gives us we solve for X. We get the cadmium concentration 5.97 times 10 to the minus 12 Moeller hydroxide. The change would be much less than 5%. So we just the used the initial concentration for the hydroxide concentration

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