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Write the balanced chemical equation describing the dissolving of each of the following sparingly solublesalts in water. Write the expression for $K_{\mathrm{sp}}$ for each process.$\begin{array}{ll}{\text { a. } \operatorname{AgIO}_{3}(s)} & {\text { c. } \mathrm{Zn}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)} \\ {\text { b. } \operatorname{Sn}(\mathrm{OH})_{2}(s)} & {\text { d. } \operatorname{BaF}_{2}(s)}\end{array}$

a. $K_{s p}=\left[A g^{+}\right]\left[I O_{3}^{-}\right]$b. $K_{s p}=\left[S n^{2+}\right]\left[O H^{-}\right]^{2}$c. $K_{s p}=\left[Z n^{2+}\right]^{3}\left[P O_{4}^{3-}\right]^{2}$d. $K_{s p}=\left[B a^{2+}\right]\left[F^{-}\right]^{2}$

Chemistry 102

Chapter 17

Equilibrium

Section 9

Solubility Equilibria

Chemical Equilibrium

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in this problem were given various solids and were asked to show their equilibrium of them dissolving in water and then determine what the cast is. So we have First, we have silver by a date, which the can I on is silver plus. So there's gonna be any quibble with a G. Crosse a Prius and I owe three minus saw agrees to um, actually, I'll just make the no Now, this first thing is always gonna be solid. The two ions to dissociate into our quest ions are always a Chris on dso or they're, like, ridiculously high anyways, So I'm gonna ignore the states at From here on out. This are equilibrium that we're dealing with. No, the equilibrium here and now our Caspi is always products overreacting. It's but we ignore solids, elected liquids. And so there's really no reactions that we're gonna consider. This is a solid nuts are only reacting. So we're gonna get the product of the concentration of a G plus times a concentration of I a date. I'll three miles. That's our answer. Such caused next one, we have 10 hydroxide, and so this is gonna be in equilibrium with 10 to plus note that since each hydroxide on is minus one and this has a neutral charge, as all solids do the 10 house have a two plus charge to counter. Act that, and we have 20 H. Miles. And so Casey's gonna be similar. But there's an important distinction you can't forget. Remember, if you have a coefficient in the equation that's going to end up being up, uh, exponential in the in our KSB expression. So we saw the concentration of sn two pots, but here we have the consolation of O. H minus square, and that's that's something you can't forget on this. Gonna be coming up often in these Caspi problems. Now, that's probably the most like, difficult thing in the sense that that's what people forget. And so our next one is that we have zinc phosphates of three sinks and Keogh for two of them, they're gonna be an equilibrium. Zinc is always a two plus charge where we can see by this coefficient there, and phosphate is always a three minus charge. And so, um, now I skipped something here. I didn't put the coefficients of front vesting. You just certainly don't want to forget. We have three sinks, as we can see here and to phosphates. And so the TSB is gonna be equal to the concentration of zinc two plus to the third. Don't forget that times the concentration of phosphate fuel for three minus squared. Don't forget that. And finally, we have one more. Um and this really won't present anything. We have barium fluoride, B A F two, which will be in equipment with very M two plus. That's how barium always is and fluoride minus two of them. Um, you can check your work by noting that we have to flow. Raj, we have to Florence on this side and note that the charges should cancel out. And so K SP is gonna be able to the concentration of barium two plus no exponential. It's It's to the first power because we technically no one there, but ignore that we have times a concentration of fluoride minus squared, and that is our answer

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