Problem

The Handbook of Chemistry and Physics (http://ope…

10:08

Problem 10 Medium Difficulty

The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives
solubilities of the following compounds in grams per 100 $\mathrm{mL}$ of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
(a) BaSiF_ $, 0.026 \mathrm{g} / 100 \mathrm{mL}$ (contains $\mathrm{SiF}_{6}^{2-} \mathrm{ions} )$
(b) $\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{4}, 1.5 \times 10^{-2} \mathrm{g} / 100 \mathrm{mL}$
(c) $\mathrm{Gd}_{2}\left(\mathrm{SO}_{4}\right)_{3}, 3.98 \mathrm{g} / 100 \mathrm{mL}$
(d) $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{PtBr}_{6}, 0.59 \mathrm{g} / 100 \mathrm{mL}$ (contains $\mathrm{PtBr}_{6}^{2-} \mathrm{ions} )$

Answer

a. $8.6 \times 10^{-7}$
b. $4.71 \times 10^{-17}$
c. $1.35 \times 10^{-4}$
d. $2.29 \times 10^{-6}$

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Chemistry 102

Chemistry

Chapter 15

Equilibria of Other Reaction Classes

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Video Transcript

So what we need to do here is right the K S P expressions. Then we need to convert the Sally abilities into Moeller so we can do the final step, which is plug in the values into the chaos. The expression to calculate ke sp. So here's my ke sp expression for this Very um um, barium species, um, coefficients are just one here, so it's just gonna be barium times s I f six. So to get to Moller of the first step is we have grams per 100 milliliters. So if we say 100 milliliters equals 1000.1 leaders, then I could just do the division and I end up with 0.26 grams per leader. So now I have the per leader part of the polarity expression. So now I just need to convert grams to moles, and we can do that with the molar mass. So this is the, um, there something erased here, but this is the molar mass of Ah, I think it's to 79. Not sure this is the molar mass of this barium barium salt B a S I f. Six. So if I just do this math. The grams cancel. I know I have one leader from up here, so that gives me 9.305 times 10 to the minus four Moeller. So this is going to be essentially the concentration of B A S i f six. And that means because the barium concentration equals the S I f six concentration. If I make this X, then K s t is X squared. So I just square the value we calculated. And that gives um, Kay s P of 8.66 times 10 to the minus seven. We just follow this protocol down through the rest of these s. So this is a looks like a Syria. My A date. Um, so when we do the Caspian expression, we have a four here. When we write out the reaction, we have a four here from the I A date on Ellen. So that's gonna be helpful when we do our K S P expression. So the first thing I want to do is convert from grams for 100 mils to Moller. So 100 mils eagle 2.1. Leaders, That gives me 0.15 grams for leader Here's the molar mass of the compound. So Grams canceled. I'm left with most for leader. So that's our polarity. So this is gonna equal the concentration of thesis Erie? Um, Ion, um, for I a day, it's gonna be four times that value. So it's gonna be four times what we calculated which would give us this value here. No, I write the chaos P expression. I have a four as our exponents from the four in the equation. So now I just plug in the values from right here into the K S P expression, and that's gonna give me this value of K S. P here. Now, with this one, there's some rounding If you ran to the end, Um, versus if you round. If you round at the end versus frowned throughout because this is just such a small number, it may be a little bit higher or lower, but it's about 4.67 times 10 to the minus 17. If I follow the same thing for this one, I write the equation. The reaction First I convert from grams per liter, two moles for leader And then I say, Okay, so we have to look at the, um we have to look at the Sochi a metre coefficients. So this gives us the Moler Sally ability of Jean de Tu s O for three. No, I have to use the coefficients to get the concentrations of each eye on. So I multiplied by two for G d three plus m multiplied by three for S O for two minus again That comes from our equation Up here. Plug into K S P expression here. The coefficients as exponents gives us 1.36 times 10 to the minus four. And for this one, it's the same thing, Um, inverting to polarity, making sure I have the concentrations of each eye on in solutions. So for ammonium, we have two multiplied by two ke sp. We have to square the ammonium because of the coefficient. And then that gives us when we do the math. 2.29 times 10 to the minus six

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