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 ion effect. Now the common ion effect decreases the solubility of a solid in 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 sulfide and placed it within pure water, it naturally will break up into its ions. So it'll break up into a barium 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 barium ion and sulfide ion. Up here, it was being thrown into pure water. Alright. The way you need to think about this is there is, basically a limit on how much of this ionic solid you can get to dissolve. In the pure water, there is none of those these ions present initially, so it's free to dissolve up to its limit based on its k s p 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 sulfide 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 callback to Le Chatelier's principle. Alright. So just keep that in mind. When an 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 when 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 2 carbonate in 0.15 molar magnesium carbonate solution. Alright. So here they're giving us the Ksp of copper 2 carbonate, so that's what we're gonna break up into ions. Here, copper 2 carbonate breaks up into copper 2 ion plus the carbonate ion. Step 1, 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 2, using initial row, place the amount given for the common ion. Alright. 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 ion? Well, in our ICE chart we're dealing with carbonate, and our solution has carbonate. They're the same ion. So that means initially we're starting out with 0.15 molar. We don't have a common ion for copper 2, so it's 0 initially. Now remember, we lose reactants to make product. So we're going to say, using the change rule, place a plus x for the products. So this is plus x plus x. Now using the equilibrium row, set up the equilibrium constant expression with k s p 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 x. Now it's this portion that we're talking about, because if we look we have 0.15 plus x. The variable x and 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, let's see, Ksp equals products over reactants. Your reactant gets cancelled out because of the solid. So we have copper 2+timescarbonate. 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 x. X equals 1.6 times 10 to the minus 9 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 5, 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 ion 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 And we can say that the solubility of an acid also decreases if the solution contains hydronium ion in the form of H plus, but also remember we can see hydronium ion in the form of H3O+. 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 solubility in grams per milliliter of chromium 3 hydroxide. You are 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 degrees Celsius. Alright. So here we have chromium 3 hydroxide solid breaking up into its ions, which are 1 chromium 3 ion plus 3 hydroxide ions. Because we're talking about Ksp, we set up an ICE chart, which stands for initial change equilibrium. Now in step 1, it says set up an ICE chart with a solid that's only reacted. Cross out the reactant side because in an ice char, we ignore solids and liquids. The reactant is a solid, so it's ignored. Using the initial row, place the amount for the common ions. Calculate Oh minus or H+ from the given pH or pOH. This is a base. Bases use pOH initially, so so we would change the pH given to us into pOH. POH equals 14 minuteus pH, so 14 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, so Oh 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 1886 times 10 to the negative 6. 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 o h minus present. We don't have any common ions for chromium 3, so this is 0. Again, this is 2.511886 times 10 to the negative 6. Now we lose reactants to make products. Using the change row, place a plus x for the products because we're making them. So plus x. There's a 3 here, so this is plus 3 x. Bring down everything, plus x, and then 2.511886 times 10 to the negative 6 plus 3 x. So step 4, using the equilibrium row, 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. If we look here, this is a real number, and it's followed by 3 plus by 3 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 3 ion times hydroxide ion, cubed, because of the coefficient of 3. Ksp is 6.7 times 10 to the minus w 1, and that equals chromium 3, which is x at equilibrium, times 2.511886 times 10 to the negative 6 cubed. We're going to 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 here will equal 4.227 times 10 to the minus 14 molar. This will be the molar solubility of my entire ionic compound. So this is for chromium 3 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 Polym3 Hydroxide per 1 liter. So here we're going to say that 1 mole of chromium 3 hydroxide, If you look at its mass, it comes out to 103.02 grams, and that's the mass of the chromium and the 3 hydroxides, all their masses together. And then finally, we need to get rid of liters, so we put 1 liter up here, and that's equal to 1,000 milliliters. 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 point 35 times 10 to the negative 15 grams per milliliter of chromium hydroxide. So this would be our final answer for this given question.
5
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)
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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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