Ksp and concentration vs soulubility

[wall1two]

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i've googled and searched but i cant find a consensus between kaplan and destroyer and chad so any help would be appreciated.

kaplan: how many moles of AgIO3(ksp= 1*10^-8) will dissolve in 1 liter of 10^-5M NaIO3 solution.

to me this means find the molar soulubility ie:

ksp = [Ag] * [IO3] = 1*10^-8 = x (aka molar soulubility) * 10^-5

so 10^-8 / 10^-5 = molar solubility right? wrong according to kaplan the answer is

2ndroot(10^-8) - 10^-5

i've never seen an expression set up like that. not in the acs or destroyer books or on chads.


also:

What is the solubility of Ca(OH)2 in 0.0860 M Ba(OH)2?

Solution:

Ba(OH)2 is a strong base so [OH¯]= 2 times 0.0860 = 0.172 M
Dissociation equation:

Ca(OH)2 (s) <===> Ca2+ (aq) + 2OH¯ (aq)
Ksp expression:

Ksp = [Ca2+] [OH¯]2
The Ksp for Ca(OH)2 is known to be 4.68 x 10¯6. We set [Ca2+] = x and [OH¯] = (0.172 + 2x). Substituting into the Ksp expression:

4.68 x 10¯6 = (x) (0.172 + 2x)2
Ignoring the "2x," we find x = 1.58 x 10¯4 M


isn't that just wrong from the start, isn't it ksp= [Ca][OH-]^2 so 4.68* 10^-6= [Ca] [ .086]^2

 

[Double Bonded]

i've googled and searched but i cant find a consensus between kaplan and destroyer and chad so any help would be appreciated.

kaplan: how many moles of AgIO3(ksp= 1*10^-8) will dissolve in 1 liter of 10^-5M NaIO3 solution.

to me this means find the molar soulubility ie:

ksp = [Ag] * [IO3] = 1*10^-8 = x (aka molar soulubility) * 10^-5

so 10^-8 / 10^-5 = molar solubility right? wrong according to kaplan the answer is

2ndroot(10^-8) - 10^-5

i've never seen an expression set up like that. not in the acs or destroyer books or on chads.

ksp=(x)(x) because the contribution from NaIO3 is negligible.

so: ksp = 1x10^-8 = x^2

so the maximum solubility/ion product is going to be x = sqrt( 1x10^-8)

but because the NaIO3 already contributed to the reaction, you just subtract out what it will contribute (Common Ion) and you're left with the maximum dissociation that AgIO3 can have. Which is

sqrt(1x10^-8) - [IO3]initially = sqrt(1x10^-8) - 1x10^-5

hope that helps...I'll get to the second one if i get time

 

[Double Bonded]

What is the solubility of Ca(OH)2 in 0.0860 M Ba(OH)2?

Solution:

Ba(OH)2 is a strong base so [OH¯]= 2 times 0.0860 = 0.172 M
Dissociation equation:

Ca(OH)2 (s) <===> Ca2+ (aq) + 2OH¯ (aq)
Ksp expression:

Ksp = [Ca2+] [OH¯]2
The Ksp for Ca(OH)2 is known to be 4.68 x 10¯6. We set [Ca2+] = x and [OH¯] = (0.172 + 2x). Substituting into the Ksp expression:

4.68 x 10¯6 = (x) (0.172 + 2x)2
Ignoring the "2x," we find x = 1.58 x 10¯4 M


isn't that just wrong from the start, isn't it ksp= [Ca][OH-]^2 so 4.68* 10^-6= [Ca] [ .086]^2

No. Ba(OH)2 dissociates into 2 moles of OH- per mol of Ba2+. That equation will look like:

Ba(OH)2 = Ba2+ + 2OH- ksp=[Ba2+][OH-]^2=(x)((2x)^2)

Initially there will be 2x.086 in the solution (or .172)

the ksp for the Ca(OH)2 equation will be:

ksp = (x)((.172+2x)^2)= 4.68x10^-6

The 2x is negligible so....ksp=(.0295X) = 4.68x10^-6

x = 1.5 x 10^-4

 

[wall1two]

thanks alot! i guess they where not looking for molar solubility but to actually calculate the moles you have to do the common ion subtraction. thanks a lot, i take this test at the end of the month and i'm pulling my hair out

 

[Kahr]

No. Ba(OH)2 dissociates into 2 moles of OH- per mol of Ba2+. That equation will look like:

Ba(OH)2 = Ba2+ + 2OH- ksp=[Ba2+][OH-]^2=(x)((2x)^2)

Initially there will be 2x.086 in the solution (or .172)

the ksp for the Ca(OH)2 equation will be:

ksp = (x)((.172+2x)^2)= 4.68x10^-6

The 2x is negligible so....ksp=(.0295X) = 4.68x10^-6

x = 1.5 x 10^-4

This is what I thought as well. 👍 I assumed the same answer for the first one as wall1two, which I guess was wrong.🙄

 

[recyrb]

ksp=(x)(x) because the contribution from NaIO3 is negligible.

so: ksp = 1x10^-8 = x^2

so the maximum solubility/ion product is going to be x = sqrt( 1x10^-8)

but because the NaIO3 already contributed to the reaction, you just subtract out what it will contribute (Common Ion) and you're left with the maximum dissociation that AgIO3 can have. Which is

sqrt(1x10^-8) - [IO3]initially = sqrt(1x10^-8) - 1x10^-5

hope that helps...I'll get to the second one if i get time

Double bonded, why in the first one is the molar solubility of NaIO3 neglegible? but in the second problem the common ion solute is significant??

 

[Double Bonded]

Double bonded, why in the first one is the molar solubility of NaIO3 neglegible? but in the second problem the common ion solute is significant??

My chem class taught it in a way that if the common ion concentration was initially low, neglect it and consider it to be zero (the 5% rule).

The second one, 2x is negligible because the 0.172 is so large relative to ksp. So you can simplify it (again the 5% rule).

In my prep for the DAT so far, I haven't seen complicated math like polynomials or something you'd usually need a calculator for.

 

[qkchen]

ok, i see the logic and understand where these numbers are coming from in the correct answer choice.

in my mind, it sounds like the original question was asking something like how much of this AgIO3 did not get dissolved, due to the common ion.

anyone want to clarify or elaborate? thanks

i've googled and searched but i cant find a consensus between kaplan and destroyer and chad so any help would be appreciated.

kaplan: how many moles of AgIO3(ksp= 1*10^-8) will dissolve in 1 liter of 10^-5M NaIO3 solution.

to me this means find the molar soulubility ie:

ksp = [Ag] * [IO3] = 1*10^-8 = x (aka molar soulubility) * 10^-5

so 10^-8 / 10^-5 = molar solubility right? wrong according to kaplan the answer is

2ndroot(10^-8) - 10^-5

i've never seen an expression set up like that. not in the acs or destroyer books or on chads.


also:

What is the solubility of Ca(OH)2 in 0.0860 M Ba(OH)2?

Solution:

Ba(OH)2 is a strong base so [OH¯]= 2 times 0.0860 = 0.172 M
Dissociation equation:

Ca(OH)2 (s) <===> Ca2+ (aq) + 2OH¯ (aq)
Ksp expression:

Ksp = [Ca2+] [OH¯]2
The Ksp for Ca(OH)2 is known to be 4.68 x 10¯6. We set [Ca2+] = x and [OH¯] = (0.172 + 2x). Substituting into the Ksp expression:

4.68 x 10¯6 = (x) (0.172 + 2x)2
Ignoring the "2x," we find x = 1.58 x 10¯4 M


isn't that just wrong from the start, isn't it ksp= [Ca][OH-]^2 so 4.68* 10^-6= [Ca] [ .086]^2

 

[wall1two]

ok, i see the logic and understand where these numbers are coming from in the correct answer choice.

in my mind, it sounds like the original question was asking something like how much of this AgIO3 did not get dissolved, due to the common ion.

anyone want to clarify or elaborate? thanks

you are exactly right, i was thrown off because the answer choices are in the unsolved format and 1*10^-5 is so small it wouldn't significantly affect a solved answer, but in the unsolved it is there.

 

[qkchen]

you mentioned that this was from kaplan. i was wondering where you saw it, was it from one of their quizzes, section tests or full length tests? appreciate your feedback

you are exactly right, i was thrown off because the answer choices are in the unsolved format and 1*10^-5 is so small it wouldn't significantly affect a solved answer, but in the unsolved it is there.

 

[wall1two]

it was from the free online tests you have to sign up for and "attend" so a bald guy can tell you to spend 1500 bucks for 3 hours. the test was too easy and truncated, a real waste of time. But it did have this ksp gem we now all got to learn from