Ksp: Common Ion Effect - Video Tutorials & Practice Problems
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1
concept
Common Ion Effect on Solubility
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2m
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With KSP, we now have to talk about the common iron effect. Now, the common iron effect decreases the solubility of a solid and a solution. Here, we're gonna say it occurs when an ionic solid dissolves in a solution containing ion or ions common to it. Now, the decrease in solubility is due to Le Chatelier's principle which remember states that the chemical reaction will shift either in the reverse direction or forward direction to decrease the disturbance to its equilibrium. If we take a look here in the first beaker, we have no common ion, meaning that I have taken this barium sulfite and placed it within pure water. It naturally will break up into its ions. So it break up into a bery ion plus a sulfite ion. Now, here we're going to say for the less soluble one, what's the difference? Well, I'm taking this barium sulfate sulfite and I'm placing it into that solution but already dissolved within the solution is some barium and some sulfite. So we're gonna write this over here. We have barium solid being thrown into a solution that already possesses be ion and sulfide ion up here was being thrown into pure water, right. The way you need to think about this is there is uh basically a limit on how much of this ionic solid you can get to dissolve in the pure water. There is none of those de ions present initially. So it's free to dissolve up to its limit based on its KSP value in the second beaker. Though some of these are already floating around, which means that this ionic solid can't basically dissolve to its maximum that it wants based on its KSP. It is limited by the fact there's already some barium and sulfite in the solution. So that means less of these would be created. So here there are common ions and because there's already some of these in the solution, the reaction will favor the reverse direction to maintain its equilibrium. And this is a call back to the Shot's principle. All right. So just keep that in mind when ionic solid breaks up in pure water, there is no common ion effect involved. The ionic solid is free to dissolve as much as it can based on its KSP value within a solution that already has some of its ions present. It's not as free to dissolve as much as it can because it's going to reach its limit sooner, right. So keep that in mind, we're talking about the common ion effect and its relation to KSP.
2
example
Ksp: Common Ion Effect Example
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3m
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Here it says, determine molar solubility of copper two carbonate in 0.15 molar magnesium carbonate solution. All right. So here they're giving us the ksp of copper two carbonate. So that's what we're gonna break up into ions. Here. Copper two carbonate breaks up into copper two ion plus the carbonate ion. Step one, we set up the ice chart with solid as the only reactant and we cross out the reactant side because remember in an ice chart, we ignore solids and liquids with an ice chart. We have initial change equilibrium step two using initial row place the amount given for the almond ion. All right. So let's go back to this question. It says that this is not breaking up in pure water. It's breaking up in a solution. That's 0.15 molar magnesium carbonate. So magnesium carbonate is made up of magnesium ion plus the carbonate ion. Each one has a concentration of 0.15 molar. Now, where's the common? Well, in our ice chart, we're dealing with carbonate and our solution has carbonate. You ever see may. So that means initially we're starting out with 0.15 molar, we don't have a common iron for co copper two. So it's zero. Initially. Now remember we lose reactants to make product. So we're gonna say using the change rope place A plus X for the products. So this is plus X plus X. Now using the equilibrium role, set up the equilibrium constant expression with KSP and solve for X, the variable X and its number can be ignored if it follows a real number. So let's fill out the rest of this ice chart. Bring down everything. This is plus X. This is 0.15 plus. Now it's this portion that we're talking about because if we look, we have 0.15 plus X, the variable X in its number can be ignored if it follows a real number. This X here follows a real number. 0.15. That means we can ignore this X and that's because in comparison to 0.15 X will be so small that it's not gonna have an impactful change on the 0.15 number. Now, here we're going to say, see KSP equals products over reactants. Your reactant gets canceled out because of the solid. So we have copper two plus times carbonate. Here KSP is 2.4 times 10 to the negative 10 equals X times 0.15. Again, we're ignoring the plus X divide both sides by 0.15. So we already have our XX equals 1.6 times 10 to the minus nine molar. This will be our answer because remember when it comes to the solubility or molar solubility of your ionic compound as a whole, that's just X. So that's what we just found. And step five, we don't need to do it because it says convert found value of X into appropriate units if necessary. Here, they want us to find molar solubility, which is typically in units of molarity. So we're already done. So this would be our final answer for the molar solubility of our ionic solid.
3
concept
Common Ion Effect: Acids & Bases
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33s
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Now, a common iron effect can also occur with acids and bases. Here, we can say that the solubility of a base decreases if the solution contains hydroxide ions already. And we can say that the solubility of an acid also decreases if the solution contains hydro ion in the form of H plus. But also remember we can see hydro ion in the form of H +30 plus. So just remember common ion effect doesn't only affect ionic solids. It can affect acids and bases as well.
4
example
Ksp: Common Ion Effect Example
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5m
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Here it says find the soil in grams per milliliter of chromium three hydroxide. You were told the KSP is 6.7 times 10 to the negative 31. If the solution is buffered at a ph of 8.4 at 25 °C. All right. So here we have chromium three hydroxide solid breaking up into its ions which are one chromium three ion plus three hydroxide ions. Because we're talking about KSP, we set up an ice chart which stands for initial change equilibrium. Now, in step one, it says set up an ice chart with the solid that only reacted cross out the reacting side because in an ice chart, we ignore solids and liquids. The reactant is a solid. So it's ignored using the initial role place the amount for the common ions calculate oh minus or H plus from the given Ph or POH. This is a base bases used POH initially. So we would change the Ph given to us into POH co equals 14 minus Ph. So four minus 8.4 which is 5.6. Now, if we know the POH, then we know how much oh minus concentration we have because oh minus which minus equals 10 to the negative POH So we're gonna say oh minus equals 10 to the negative 5.6 which comes out to be 2.51 one 886 times 10 to the negative six. And we're going to say that that amount of oh minus represents the initial amount of our oh minus. Within this question, we're being placed in a solution with that Ph which means it already has that much oh minus present. We don't have any common onions for chromium three. So this is zero. Again, this is 2.511886 times 10 to the negative six. Now we lose reactions to make products using the change role place A plus X for the products because we're making them. So plus X, there's a three here. So this is plus three X bring down everything plus X and then 2.511886 times 10 to the mega six plus three X. So step four, using the equilibrium row, set up the equilibrium cosmic expression with KSP and solve for X. The very X and its number can be um ignored if it follows a real number. If we look here, this is a real number and it's followed by three plus by three X because of that, we can ignore this portion. Now, here we'd have to convert the found value of X into appropriate units if necessary. So let's just solve for X initially. So KSP equals products. So it's chromium three ion times hydroxide ion cubed because of the coefficient of three KSP is 6.7 times 10 to the minus 31. And that equals chromium three which is X at equilibrium times 2.511886 times 10 to the negative six QED. We're gonna say here this is 6.7 times 10 to the minus 31 equals X times. So I'm just gonna take the Q of this value when I do that, that gives me 1.58 times 10 to the negative 17 divide that out from both sides. Now, when I do that, that'll give me X X zero equal 4.227 times 10 to the minus 14 molar. This will be the more solubility of my entire ionic compound. So this is for chromium three hydroxide. Now, they don't want it in molarity, which is moles per liter. They want the solubility in grams per milliliter. So we're gonna do some converting here. Now, remember molarity itself is moles over liters. So this number is equivalent to 4.227 times 10 to the negative 14 moles of chromium three hydroxide for 1 L. So here we're going to say that one mole of chromium three hydroxide. If you look at its mass, it comes out to 103.02 g. And that's the mass of the chromium and the three hydroxides, all their masses together. And then finally, we need to get rid of liters. So we put 1 L up here and that's equal to 1000 mL. Here, liters cancel out. And what we have at the end will be grams per milliliter. So when I work that out, I get 4.35 times 10 to the negative 15 g per milliliter of chromium, three hydroxide. So this would be our final answer for this given question.
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Problem
Problem
Which of the following compounds will become more soluble in basic solution?
a) PbF2 (s)
b) ZnCl2 (s)
c) Al(OH)3 (s)
d) MgCO3 (s)
A
PbF2 (s)
B
ZnCl2 (s)
C
Al(OH)3 (s)
D
MgCO3 (s)
E
a) and b)
F
b) and c)
G
b) and d)
6
Problem
Problem
A solution of Ba(OH)2 has a Ksp of 5.0 x 10−3.
i) Determine the pH of this solution.
ii) Determine the pH if Ba(OH)2 was added to a solution containing 3.2 M of BaF2 and 0.94 M of Al(OH)3.
A
i. 2.30 ii. 14.03
B
i. 0.67 ii. 13.55
C
i. 13.33 ii. 14.45
D
i. 2.30 ii. 13.65
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