Solubilty Product Constant - Video Tutorials & Practice Problems

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TheÂ solubility product constantÂ represents the equilibrium value for an ionic solid and its ions in a given solution.Â

Solubility Product Constant

1

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Solubility Product Constant

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So here we're gonna say that saul ability represents the maximum amount of solute that could successfully dissolve in a solvent. Remember when a saw you successfully dissolved within a solvent, we create a homogeneous mixture known as a solution. Now we're gonna say associated with any solid and the ionic solid is a Ks. P. Value. This K. S. P. Value is our solid ability product constant. We're gonna say here, the larger your sociability product constant than the more soluble an ionic solid is in a solvent. So the more of it breaks up. And then we're gonna say here, the smaller the psy ability product constant and the less soluble an ionic compound becomes within a solvent. So just remember if we're looking at the suitability of any ionic compound associated with it is a Ks. P. Value. Now that we've gotten that out of the way, click onto the next video and see how we look at example one. When we're asked to figure out what are some liability, product constant is for a hypothetical ionic compound when given its molars liability

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example

Solubility Product Constant

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So here we're told we have a hypothetical compound with the formula mx three and it has a solar cell viability of 30.562 molars. Were asked, what is the value of the cell viability product constant for this hypothetical ionic compound. Now with K. S. P. Here we're dealing with our ionic compound. It's gonna break up into its ions. The metal Which is M breaks off this little three here came from the metal so its original charge was three plus. Remember, ions in solution are Aquarius plus there are three of the X. Negative ion. So here we have our equilibrium equation with K. S. P. We're talking about how it forms an equilibrium and with equilibrium will set up an ice chart. So we have initial change equilibrium. Remember we ignore solids and liquids within an ice chart? They're both breaking the whole thing is breaking up in pure water so initially our ions are zero and zero but their products or their being formed. So the change is going to be plus X. For this, this is a three here. So this is plus three X. Bring down everything. So plus X plus three X. Now K. S. P. Is an equilibrium constant. And like all of the equilibrium constants, it equals products. Overreact ints. But here, when we're dealing with a saturated solution, we're gonna say here that we have our ionic compound is a solid so it's going to be ignored. Remember we ignore solids and liquids. So que sp will just equal products. So in this case it will equal M three plus times X minus the three here makes this three X. But it also cubes. It it becomes the power we're gonna plug in. What we know at equilibrium for each one of these ions. So at equilibrium M three plus is X. And X minus is three X. So plug it in. So X times three X. Which is cute. So that's gonna be X times three cubed is three times three times three which is 27 then this is X cubed. Multiplied together, gives us 27 X. To the fourth. So that's what K. S. P. Equals. Now realize here when they give us the saw ability of the ionic compound that saw ability itself represents X. So all we're gonna do now is plug that value in for X. So we'll write it over here. So K. S. P. Equals 27 X. To the fourth. So that's 27 times 270.562. Which is gonna be to the fourth. What you're gonna do here is take that number to the fourth then multiply by 27. So 27 is gonna multiply 9.9754 times 10 to the negative 10 Multiply that by 27. So that comes out to 2.69 times 10 to the -8. And like other equilibrium constant K. S. P. Has no values. So it's just that number which gives us option D. So just realize here when they're telling us the Mueller saw liability, they're giving us what X equals for the entire ionic compound. Using that is key to determine your K. S. P. Now that you've seen example one take a look and see if you can do example to hear. We're asked to figure out the highest moller saw the ability of each ionic compound given. So here we have to determine what the X. Variable will be from our working example one, you just have to look at things backwards, moving in the opposite direction to help us to isolate X for each situation. And from that you'll be able to see which one has the highest X. Value and therefore the highest moller saw ability.

3

example

Solubility Product Constant

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here it states which of the following compounds will have the highest moller saw inability in pure water. Now to determine which one has the largest smaller liability, What we should do first is determine how many ions does each ionic compound break up into. So first we have cobalt to hydroxide that will break up into when we work it out, we're gonna have one cobalt 2 ion Plus two hydroxide ions. So that's three ions total that we have here. Next we have strontium phosphate. So that's gonna break up into work it out three strong team ions Plus two phosphate ions. So so far we have three ions here. Five ions here. Next we have led to chloride that breaks up into lead to ion plus two chloride ions. So that's three ions so far. Next we have silver cyanide Which breaks up into one silver ion Plus one Cyanide Ion. So what do we have so far? Two ions there and then finally we have lead to sulfate, which breaks up into lead to ion plus sulfate ion. Alright, so let's look at the ones that have a number of ions in common. So we have first Cobalt two hydroxide and lead to chloride. They both have the same number of ions. So now we can look at their K. S. P. And determine which one is more. Um We'll have the higher Mueller sought ability. So here lead to chloride has a higher Ks. P value because it's 10 to the negative five. So that means that it's more soluble than cobalt to hydroxide would be because it's more soluble. That means going to have a larger moller Salyer bility. So automatically we know that C. Is greater than a. Next we look at silver cyanide and lead to sulfate because they both break up into two ions. Now comparing their Ks piece here, we're going to say that this is to the negative eight. Whereas this one's to the negative 17. So let two sulfate is more soluble. Therefore it's gonna break up more easily into its ions meaning it will have a higher moller cell viability. So this is out. So we have left are strontium phosphate led to chloride and lead to sulfate here. We can't just simply look at them and look at their K. S. P. S. Because they break up into a different number of ions. In this case we'll have to determine what our solar cell viability or X. Is for each ionic compound. So we'll start out first with strontium phosphate and see how it's done. Alright so we're gonna take this and work it out. Remember we're dealing with an ionic solid. Breaking up since it's an ionic solid. This is out these two are products that are being made so this is zero and zero. There's three here. So this is plus three X. There's a two here. So this is plus two X. Plus three X. Plus two X. K. S. P. Here will equal strontium ion and it's cute because of the three as its coefficient times phosphate ion squared because also it's coefficient. So we're gonna say here that comes out to three X. Which is gonna be cubed and two X. Which is going to be squared. So that's 27 X. To the third times four X. To the two which comes out to 108 X. To the fifth. So my K. Sp here which is 4.0 times 10 to the negative 28 equals 108 X. To the fifth. Right? So At this point divide both sides here by 108. So when we do that we're gonna say now X to the 5th Equals 3.70 times 10 to the -30. Take the 5th route of both sides. So X equals 1.30 times 10 to the negative six moller. This number here represents the moller saw the ability of the ionic compound itself. All we have to do now is do it also for lead to chloride and lead to sulfate. So we can get rid of this choice because we know that's not gonna be an answer. We're gonna bring this portion down. Mhm Over here. Alright so initial change equilibrium X. So this is zero plus zero plus X. There's a two here. So plus two X plus X plus two X. K. S. P equals PB two plus times cl minus squared equals X times two X squared which comes out four X. To the to the three. And this equals 1.60 times 10 to the -5, Divide both sides here by four. So that's gonna give me now. X cubed equals 4.0 Times 10 to the -6. Take the cube root of both sides now And that's gonna give me x equals .0159 molar. So so far we know that lead to chloride will be better than a strong tim phosphate because it has a larger X. Value. So it's more of a liability is greater here. This is initial change equilibrium again 00. This is plus X. And plus X plus X. And plus X. K. S. P now equals PB two plus times S. 042 minus. So that's X squared K. S. P. is 1.8. 2 times 10 to the -8. Take the square root of both sides here just to isolate X. So X. Here for the final one will equal 1.35 times 10 to the negative four moller. So we can see here that the answer has to be option C. Because they gave us the largest X. Value. So it has the largest moller soy ability. So remember for a question like this, a strategy is to first determine how many ions they each break up into then group the ionic compounds that break up into the same number of ions when you do that, you compare their K. S. P. S to one another, the one with the larger K. S. P. In that case would be more soluble and therefore have a higher moller saw the ability. Once you've done that, all you'll be left with is ionic compounds that each break up into different number of ions. In each case, you'll have to calculate what X. Is at the end and by comparing, you'll know which one is the most soluble overall because it has the largest Mueller sought ability.

Solubility Product Constant Calculations

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example

Solubility Product Constant Calculations 1

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So here we're told that lantern um three hydroxide has a K. S. P. Of 2.0 times 10 to the negative 21. It says how many grams of lantern um three hydroxide are dissolved as hydroxide ions and 2.5 liters of a saturated solution of lantern um three hydroxide. Alright so we're dealing with the K. S. P. Of this compound. If we're dealing with K. S. P. That means we're dealing with the solid form of this compound and we look to see how it dissolves in water. When it dissolves in water it establishes an equilibrium. So we're gonna have one lantern um three ion ions when dissolved in water. Aquarius plus three hydroxide ions. Here we're gonna set up an ice chart when dealing with K. S. P. So we have initial change equilibrium with a nice chart. We ignore solids and liquids. So here my solid will be ignored. Products are being formed over time. So this will be 00 plus X. Because this is a three. This is plus three X plus X plus X plus three X. Now we're gonna say K. Sp which is your cell viability. Product constant equals products. Overreacting is just like all of the equilibrium constants. But here for this equation are reacting is a solid. So we're going to ignore it. So que sp here will just equal products. So it equals lithium, not lithium um lantern um three ion times O. H minus because again this coefficient is three. Now this is gonna be to the third power. So that coefficient makes it three X. But also makes it raised to that power. K. S. P. Again is 2.0 times 10 to the negative 21 at equilibrium. L. A. Three plus is X. O. H minus is three X. So that's X times so three cubed is three times three times three which is 27. And then that's X cubed. So this comes out to 27 X. To the fourth equals 2.0 times 10 to the negative 21. So here we need to isolate X here, Divide both sides here by 27. When we do that here, that's gonna give me seven point 40741 times 10 to the negative 23 Equals X to the 4th. Now we need to just isolate X. Not X. To the fourth. So I'm gonna take the fourth root of both sides. So forth route. So for some of you you may see this button on your calculator, you'll see an X. And then the square root symbol. What you would do is you punch in four first. Then you hit that number brackets, plug in this value that we have closed brackets and that give us X. For those of you don't have that you may have to hit. Um You would have to plug this number in still in parentheses and you may see this fun you may see carrot button that points up like that or you might see why did the X. So you have one of those two buttons. And then you do parentheses 1/4 because cube four through it is the same thing as to the 1/4 power. And that would give you the same X. As an answer. So that will give us X equals 2.934 times 10 to the negative six molar. So that is what our X. Variable is equal to anytime. We find X. X. Here represents the salt ability of my ionic compound. If we go back to the original question, it's asking us to find grams how many grams of lantern um three hydroxide are dissolved as hydroxide ions. We're looking for grams of hydroxide ions. The wording is a little bit tricky but that's what we're looking for. So if we need to find the grams of hydroxide ions, we need to first figure out its concentration at equilibrium. Hydroxide is equal to three x. So this X. That we just found, plug it into here to find the concentration of O h minus. So again O h minus equals three exit equilibrium. So that's three times the answer that we just got. So h minus here equals 8.80 Times 10 to the -6 Molar. Remember that moles Equals leaders times more clarity. We're told that we have 2.5 L. So multiply that by the polarity. We just isolated for hydroxide ions and that will give us the moles of hydroxide ions. That comes out to 2.20 times 10 to the negative five moles of hydroxide ions. All we have to do now is just change those moles into grams. So one mole of hydroxide ions, We have one oxygen, one hydrogen. So looking at your periodic table, the masses are 15.9994 g and 1.794 g, Add them together, gives us 17.0073g roughly of hydroxide ions moles cancel out. And we'll get grams at the end which comes out to three 74 times 10 to the -4 g of hydroxide ion. So as you can see there's a great deal of work that's required in terms of answering this question, Read the question very carefully here. We're not looking for grams of the entire ionic compound. We're looking for grams of length, lantern um three hydroxide in the form of hydroxide ions. So we're actually looking for the grams of hydroxide ions in this question. Any time we find X. X. Gives us the solid ability or concentration of our ionic compound as a whole. If we want the concentration of the particular ions, we have to look at the ice chart at equilibrium. This ion was just equal to X. So it'd be the same exact number. But for hydroxide ions at equilibrium it equals three X. So it'd be three times what we found for X. To get the concentration of hydroxide ion from there, we can change concentration two moles by using leaders and then change those moles into grams. Now that you've seen this example move on ahead to example to see if you can approach this question. If you can't don't worry, just come back and see how I tackle this example to question.

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example

Solubility Product Constant Calculations 1

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So here it says find the ph of a saturated solution of aluminum hydroxide. We're told that the K. S. P. Of aluminum hydroxide is 1.9 times 10 to the negative 10. Alright so we know that we're dealing with K. S. P. So we're dealing with this ionic solid. And we're discussing how it breaks up into ions when thrown into solution. So breaks up into aluminum ion Plus three hydroxide ions. Hopefully by following the example up above you're able to get this far in terms of this problem. The steps are very similar to what we saw up above. Except now we're being asked to find ph now here we're gonna have initial change equilibrium again. Remember solids and liquids are ignored within our ice chart. We're gonna say here initially this is zero and this is zero. This is plus X. Just like up above we have +30. H minus is here. So this is three X. So plus X plus three X. K. S. P equals products. Overreact ints. But we ignore the reactant because it's a solid. So it equals aluminum times O. H minus. It's gonna be cubed because of the three as a coefficient K. S. P. Is 1.9 times 10 to the negative 10 X Times three X. Which will be cubed. So just like up above we have 27 X. To the fourth again Equals 1.9 times 10 to the negative 10, Divide both sides here by 27. And we'll get what X. Fourth X. To the fourth equals. So X. To the fourth equals 7.3704 times 10 to the negative 12. Again like up above we're gonna take the fourth root. Or you could do it to the 1/4 power to find X. X. equals 1.629 times 10 to the -3 Moeller. This represents the concentration of my entire ionic compound. But remember we're looking for ph here. Now in our equation we don't have a choice 30 plus anywhere. But what we have instead is oh H minus. If I know H minus is concentration I can find my P. O. H. And from there I can find ph so let's figure out what the concentration of O. H minus is at equilibrium. O. H minus equals three X. So h minus equals three X. So that's three times 1.629 times 10 to the minus three. So that's gonna come out to 4.886 times 10 to the -3 moller. Now that we have the polarity, we can figure out what my ph is. So first we're gonna take the negative log of this number to find P. O. H. Which is the negative log of O. H minus. So that comes out to 2.31. Now that we have P. O. H. We know ph because ph equals 14 minus P. O. H. So that comes out to 11.69 as my ph for this solution. So if you are able to follow along with the example up above this question was similar up to a point once we found the concentration of O H minus in this example, we didn't need to figure out its grams or it's moles. Instead, we'd use those concentrations that we found to figure out our p O H from p O. H, transforming into p H. And we have our answer at the end. So, following all the normal conventions that we saw up above, we see that our ph is equal to 11.69.

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example

Solubility Product Constant Calculations

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So here it says calculate the molar saw the ability of iron to carbonate in a solution of 0.167 moller of sodium carbonate here we're told that the K. S. P. For iron to carbonate is 2.1 times 10 to the negative 11. Alright, so we're dealing with iron to carbonate and were given its K. S. P. Which means that we know we're dealing with the solid form of it. We look at how it breaks up into ions. Now realize up to this point we're normally seeing these ionic compounds breaking up in pure water. But in this case it's breaking up in a solution that is comprised of sodium carbonate, realize sodium carbonate breaks up into sodium ion and carbonate ion. And the carbonate ion is what's important because that carbonate ion is the same carbonate ion here and we're gonna say here because we're breaking up in a solution that already possesses carbonate ion, we're dealing with the common ion effect. So with the common iron effect, this is a solid organ, ignore. There's no common iron of this, this zero with the common iron effect because this carbonate, which the solution is comprised of, is the same as this ion here. The initial concentration will be .00167. Then they're both products or they're both being made. So this is plus X plus X plus X. And then 0.167 plus X. Now with K. S. P. Remember we're concerned with the product because the reaction here is a solid, it's ignored. K. S. P. Equals F. e. two plus times C. 0. 3 2 -. So here K. S. P. Is 2.1 times 10 to the negative 11 equals X. Times Now with K. S. P. We're gonna say because K. S. P. Values are normally so low for the sake ability of these ionic compounds. That means that we can ignore our X. Variable as long as there is a number That is in front of it that precedes it. So we're gonna have .00167 here we're looking for the moller saw the ability of iron to carbonate. So that just means we're looking for X. So divide both sides here by .00167. And when we do that that's gonna give us our X. Variable which is our molars liability for the entire ionic compound that comes out to 1.26 times 10 to the negative eight moller. So remember with the common i in effect we're no longer breaking up our ionic compound in pure water. We're breaking up in a solution that possesses a common ion to one of the ions in are ionic compound. So taking this to heart, attempt to do practice question one. If you get stuck don't worry just come back and take a look at how I approach that same exact practice question and how we get the answer that we're looking for

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Problem

Problem

What is the molar solubility of Fe(OH) _{3} (s) in a solution that is buffered at pH 3.50 at 25Â°C? The Ksp of Fe(OH)_{3} is 6.3 x 10^{â€“38} at 25Â°C.

A

1.99x10^{-27} M

B

1.99x10^{-34} M

C

6.64x10^{-28} M

D

1.99x10^{-6} M

E

1.25x10^{-9} M

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Problem

Problem

Arsenic trisulfide (As_{2}S_{3}) occurs naturally as the orange-yellow colored mineral orpiment. As_{2}S_{3} is a highly insoluble substance with a Ksp value of 2.90Ã—10^{âˆ’72}. Calculate the solubility of As_{2}S_{3} in g/100mL. (Molar mass of As_{2}S_{3} = 246.04 g/mol)

A

2.12Ã—10^{âˆ’17} g/100 mL

B

1.87Ã—10^{âˆ’13} g/100 mL

C

6.25Ã—10^{âˆ’15} g/100 mL

D

4.75Ã—10^{âˆ’14} g/100 mL

The Reaction Quotient & Precipitation

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Reaction Quotient & Precipitation

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So here we're talking about the reaction quotient Q. We're gonna say here that the reaction quotient Q. Is used to determine if our chemical reaction is at equilibrium when we compare it to K. Or if we produce a precipitator solid and that's when we compare it to K. S. P. Remember regular Ks. Your equilibrium constant. K. Sp is also an equilibrium constant, but K. S. P. Is an equilibrium constant for ionic solids. Now here we're gonna say if the reaction quotient Q is equal to the equilibrium constant K. Then our reaction is at equilibrium. For example, here we have silver chloride as our solid, it breaks up into silver ion and chloride ion here were given the concentrations initially of silver and chloride as 1.3 times 10 to negative five molar. Now Q is just like K. It equals products. Overreact ints. And just like our equilibrium constant, K. S. P. We ignore solids and liquids, so Q. Equals just products. So it equals silver ion times chloride ion plugging those values in gives us our Q. Value here are Q. Is 1.77 times 10 to negative 10. Just like K. S. P. Is 1.77 times 10 to the negative 10. They're equal to one another. So we can say because Q and K equal each other here we're at equilibrium in terms of shifting of our chemical reaction, there would be no shifting of our chemical reaction because our chemical reaction is at equilibrium. Now later on when we look up new values for Q. If Q happens to be different from K, this will determine the direction our chemical reaction will shift in order to re attain equilibrium conditions. The shifting can have an effect on the amount of reactant or product that's used or created. Now, click onto the next video and see what happens when our Q value is different from r. K value or k. S. P value.

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Reaction Quotient & Precipitation

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So here we continue our discussion of the comparison of the reaction quotient Q to any type of equilibrium constant, including K. S. P. Now in these next two scenarios, they will not be equal to one another because Q and K will not be equal to one another. This will cause my chemical reaction to shift either in the four directions or the reverse direction to reestablish equilibrium. Now, in this first example here were given new concentrations for ions. So here we have 40.250 molar for each one. And when we plug them into the equilibrium expression for Q, We can see that the value of Q comes out to 6.25 times 10 to the -2. In this case we're gonna see that Q. Is bigger than our K. S. P. Which is still the same value. Remember anyway this K. S. P. Or any type of K would change as if there was a change in temperature. So R K. S. P is remaining the same because temperature is being held constant. But the concentration of my ions are changing which will have a direct impact on the value of Q. So here Q. Is larger than K. S. P. So if Q. Is larger than K. S. P. Or K in general, then our reaction will shift. So on terms of our number line Q will always shift to wherever K. Is. So here Q will shift this way to get the K. And if it shifts that way to get to K, then our chemical reaction will also shift as a whole to get to equilibrium. Remember wherever you shift that direction will cause an increase. So here we're shifting towards the react inside. So the reacting side would be increasing. If the reacting side is increasing, that means our product side is decreasing. So we're gonna say if Q. Is larger than K, then our chemical reaction will shift to the reactant side in this case because we're shifting towards the reacting which happens to be a solid. That means that the precipitate will increase. The amount of solid will increase. Now here, in this case we have new concentrations yet again for the ions again temperatures not being affected. So our equilibrium constant K. S. P remains the same. So in this case we have these new concentrations which gives us a new Q value which is 5.0 times 10. 5.6 times 10 to the negative 24. We can see now that that value is less than K. S. P. So if Q. Is smaller than K. S. P. Or K in general. So now he was no longer here. Instead you will be over here, it's gonna be smaller. So we're gonna have to shift this way. Now we're gonna shift that way in terms of the number line. And in terms of chemical reaction, remember wherever we're heading towards that side is increasing. So this side would increase this side here would decrease. So our chemical reaction will shift to the product side. So that means that the cell ability of my ionic compound is increasing because I'm making more ions and the amount of precipitator solid will decrease. So just remember Q. Our reaction quotient is just used as a way of determining. Are we at equilibrium When Q equals K. S. P. We're at equilibrium. We will say that our chemical reaction or actually our solution is a saturated solution with no real dominant movement to the reacting side of the product side. Where at equilibrium everything is balanced. So we represent a saturated solution. Then if our Q happens to be larger than R. K. S. P. That means we have an excess of ion. So that's why the reaction is shifting in the reverse direction towards the reactant to recreate some more precipitate. We'd say that this type of process happens within a super saturated solution. And then finally when Q. Is less than K. Sp, we can continue to break down our ionic solid to produce more ions. So when Q. Is less than K. S. P. We'd say that we are an unsaturated solution now that we've explored the connection between our reaction question Q. And K. S. P. Will be more prepared when we take a look at questions dealing with more calculation based questions on the formation or lack thereof, of precipitates

When Q is equal to K_{sp }the solution is saturated and there is no shifting in the equilibrium position.

When Q is greater than K_{sp} the solution is supersaturated and the reaction shifts in reverse to form more precipitate.

When Q is less than K_{sp} the solution is unsaturated and the reaction shifts forward to form more ions.

The Reaction Quotient & Precipitation Calculations

11

example

Reaction Quotient & Precipitation Calculations 1

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here it asks will it precipitate form when 0.150 liters of 0.100 Moeller led to acetate and 0.100 liters of 0.20 molar sodium chloride are mixed here, we're told that the K. S. P. Value of lead to chloride is 1.2 times 10 to the negative five. Alright, we're given the K. S. P. Of lead to chloride. So that's our ionic compound of interest. So we're gonna take that ionic compound and bring it down. So it's our solid. And we show how it dissociates into its ions when it's thrown into our solution. So that creates led to ion acquis plus two chloride ions. And here we have initial change equilibrium realize here at this point is um we have this led to chloride forming. And what's what's happening here is we have a solution made up of lead to chlor led to acetate and sodium chloride. They're breaking up into ions and those ions are combining together to help form this solid here and all we're doing is showing its equilibrium equation. So this led to chloride exists in a solution made up of these four particular ions. It's not being formed in pure water. And if you look those four ions from those compounds, we have lead to ion in it and we care about that because that's part of my equilibrium equation. And then we have chloride ion here which is also part of my equation. So here we're dealing with double kalmunai in effect these amounts that we're gonna find for both the lead to close. Well the lead to ion and the chloride ion will be used to help us determine what Q. Will be. So we have to figure out what the concentration of this led to ion is and the concentration of this chloride ion is. Remember leaders of polarity of means multiply. Remember that moles equals leaders times more clarity. So we take .150 L Multiplied by the 0.100 Moeller of lead to acetate. That's gonna give me .0150 moles of lead to acetate. Realize here that we don't want the lead to ascertain we just want the Mosul led to ion because that is one of our common ions. So we're gonna stay here for every one mole of the entire compound. We have exactly one mole of lead to ion involved. Now we need to determine its concentration or its polarity. Realize here that we have 0.150 liters of one compound mixing with exactly 10.100 liters of another compound. Their combined volume forms the new overall volume of the solution. So for the molds we just found we're gonna divided by the total volume. So that's the volume from the two mixtures. The two compounds doing that will give you my new concentration for like two ions which comes out to 20.60 moller. Now we have to do the same thing and figure out the new concentration of chloride ions. So we have 0.100 liters of 0.20 molar sodium chloride again. Remember that Leaders. Times more clarity will give me moles. So bring that down. We don't want the moles of sodium chloride, we just want the moles of chloride ion itself. So for every one mole of sodium chloride we can see that there's exactly one chloride in the formula. So that's 0.2 moles of chloride ions. Again we want to find its new concentration. So we divided by the total volume used. So that's 0.150 liters Plus .100 L on the bottom Together. All of that gives me a polarity of .080 moller for my chloride ions. So take that concentration and plug it in. So now we have initial concentrations for both. So this would be plus X. Plus two X .060 plus x .080 Plus two x. Because we have initial concentrations for both ions. This is going to help me figure out what my Q. Value will be. So we're gonna say here Q equals um just products because are reacting is a solid. It's ignored. So that's PB two plus times cl minus squared. So it's gonna be .060 times .080 Squared. When we plug that in. That gives me 3.84 times 10 to the -4 as my Q value. Now remember what we said in the previous videos is at this point when we find Q. We can compare it to our K. S. P. Using a number line that will dictate which direction my chemical reaction will shift. So here's our number line here, we'll put K. S. P in the middle. So we're told that K. S. P. Have led to chloride is 1.2 times 10 to the negative five. We can see that Q. Is this number 3.84 times 10 to the negative four. Here it's to the negative four. So it's a larger value. So Q will be over here. All we say now at this point is because K and K S. P and Q are not equal to each other where you are not at equilibrium, Q will shift in the direction needed to get to K. S. P. So Q will shift this way to get to K. S. P. Remember whichever direction you shift in to get to K. Is the same direction your reaction will shift to as well. So here my chemical reaction would have to shift also this way, wherever you're shifting will be increasing in amount. So my reactant here is increasing because we're shifting towards the reactant and this side here is decreasing. So we're saying here that my precipitate, which is my solid is increasing in amount. So we would say yes a precipitate does form. Remember when Q is greater than K. S. P. A precipitate will begin to form because our solution represents a super saturated solution and to get back to equilibrium, some of the extra dissolved solute has to re crystallize in the solution itself. So again here, because Q is greater than K S P R reaction moves in the reverse direction towards reactant, thereby creating more solid. Now that you've seen this example. Look to see if you can tackle example too. If you get stuck or lost. Don't worry, just come back and take a look at how we approach example too.

12

example

Reaction Quotient & Precipitation Calculations 1

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So here's the question states what is the minimum ph at which iron to hydroxide will precipitate if the solution has a concentration for iron to ion equal to 0.583 molar here, we're told that the K. S. P. Of iron to hydroxide is 4.87 times 10 to the negative 17. Now this question here, although it's asking for precipitation, it's really piggy, backing off the idea of just the simple common iron effect of iron to ion. So we're gonna write the ionic compound we know here it's gonna form an equilibrium with its ions. So it's F. E. Two plus plus two hydroxide ions. So we're gonna have initial change equilibrium. We ignore the solid Here. This iron to hydroxide compound is not breaking up in pure water. But in fact the solution already comprised of iron to ion. So we have .0583 moller iron to ion. So that their represents the initial concentration. Here we're gonna see the initial concentration of hydroxide is zero. Since we don't have that common ion at all present in our a solution. So this is gonna be plus X. Plus two X 20.583 plus X plus two X. Here we're dealing with K. S. P. We're not dealing with Q. We dealt with you in the previous question because we had initial amounts for both of my products initially here because we only have one of them. It's just a simple common in effect that also encompasses a question dealing with a precipitate. So here K. S. P. Equals F. E. Two plus times O. H. Minus. And it'll be squared equal. So remember when we have an actual value we can ignore the variable. That's because the variable is gonna be such a small number. It's not gonna have a heavy impact on the Value we have here. So that's gonna be .0583 Times two X. Which is going to be squared And K. S. P. is 4.87 times 10 to the -17. Alright so here we're gonna have 0.583 so two X. Squared. So to become squared. So that's gonna become four and that's X. Squared. Then here we're gonna have we're gonna have this times four which is gonna give me 40.2332 times X squared, Divide both sides by .2332. So X squared is gonna equal to 0.8834 times 10 to the negative 16. We want the square root of both sides now so X. Here at this point equals 1.445 times 10 to the negative eight moller. Now they're asking us for the minimum ph realize here that ph is just the negative log of H plus concentration. What we have. That's similar to that is O. H minus. So we're gonna do here is we're actually gonna figure out what the concentration of O. H minus is first. So at equilibrium O. H minus equals to X. So plug in the X. That we just found. So that's gonna give me a concentration of 2.89 times 10 to the negative eight moller as the concentration of my hydroxide ion. If we know the concentration of hydroxide ion then we know P. O. H. By default because it's equal to the negative log of O. H minus. So that's negative log of 2.89 times 10 to the -8. So that's gonna give me 7.54. So that's my P. O. H. I don't want P. O. H. I want P. H. P. H equals 14 minus P. O. H. So it's 14 -7.54 which comes out to 6.46. So that that represents the minimum minimum ph at which a precipitate of iron to hydroxide can become a precipitate. Now realize here again we didn't use que in this question because we didn't have the initial concentrations of both of my products because of that this is just a common i in effect question to help us figure out the concentration in which we can get O. H minus and from that we can figure out ph eventually