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### Topic: solubility solve for ksp  (Read 18803 times)

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#### plasticfood

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##### solubility solve for ksp
« on: November 12, 2011, 11:30:30 PM »
the solubility of AgCl in water is 3.97 * 10^-6 M @ 0 Celsius and 1.91 * 10^-4 M @ 100 Celsius. Calculate the Ksp for AgCl for both temperatures.

So is ksp = [Ag+][Cl-]? the first ksp = (3.97 * 10^-6)(3.97 * 10^-6)?

#### UG

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##### Re: solubility solve for ksp
« Reply #1 on: November 12, 2011, 11:51:05 PM »
So is ksp = [Ag+][Cl-]? the first ksp = (3.97 * 10^-6)(3.97 * 10^-6)?
Yes exactly

#### plasticfood

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##### Re: solubility solve for ksp
« Reply #2 on: November 13, 2011, 06:41:53 PM »
and one last question, i need to calculate delta S from the same problem above. it says delta S is equal to b in the equation y = mx + b. i know how to solve for the slope, but how do i find the y-intercept?

#### UG

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##### Re: solubility solve for ksp
« Reply #3 on: November 13, 2011, 06:53:17 PM »
and one last question, i need to calculate delta S from the same problem above. it says delta S is equal to b in the equation y = mx + b. i know how to solve for the slope, but how do i find the y-intercept?
If there isn't one, you'll probably need to give us a bit more information on the question.

#### plasticfood

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##### Re: solubility solve for ksp
« Reply #4 on: November 13, 2011, 07:09:05 PM »
i'm not given a graph, but i think i'm still using the same info as the first question: the solubility of AgCl in water is 3.97 * 10^-6 M @ 0 Celsius and 1.91 * 10^-4 M @ 100 Celsius.

so do i graph ln ksp vs. 1/T?? there's another question that ask to solve for delta H, so i use the slope formula and solve for m, so then i can use m = -delta H / 8.314. but how do i find the y-intercept to solve for delta S?

#### UG

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##### Re: solubility solve for ksp
« Reply #5 on: November 13, 2011, 07:17:41 PM »
Hi plasticfood, please be specific in what the question asked. At the moment I am slightly lost (although it sounds like you calculated the Gibbs free energy at each temperature )

#### plasticfood

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##### Re: solubility solve for ksp
« Reply #6 on: November 13, 2011, 08:08:32 PM »
ok so this is the worksheet: http://imgur.com/0HyaX

i am stuck on #3, but now i'm not quite sure if i did #2 right either.
i am just confused b/c the way i figure out delta H and S from the lab is looking at the equation of the line, but now i don't have the equation i guess??

#### UG

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##### Re: solubility solve for ksp
« Reply #7 on: November 13, 2011, 08:17:31 PM »
Ok, I see now
So what you need to do is firstly calculate the Gibbs free energy change of the reaction at each temperature.
The formula is ΔG = -RT ln Ksp so you should get two values for ΔG, one at 0oC and one at 100oC (but remember in the formula T is in Kelvin). Then the formula you use to plot your line is ΔG = ΔH - TΔS
So you need to plot ΔG vs. T (in Kelvin). The slope of this graph will be -ΔS and the y-intercept will be ΔH.
« Last Edit: November 13, 2011, 09:15:52 PM by UG »

#### plasticfood

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##### Re: solubility solve for ksp
« Reply #8 on: November 13, 2011, 08:50:10 PM »
the first delta G is -3.56e-8 and the second is -1.1317e-4. i've tried plotting this in excel, but the graph is not coming out right. i know to how to manually solve for the slope of the line, but how do i manually solve for the y-intercept?

#### UG

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##### Re: solubility solve for ksp
« Reply #9 on: November 13, 2011, 08:56:29 PM »
the first delta G is -3.56e-8 and the second is -1.1317e-4. i've tried plotting this in excel, but the graph is not coming out right. i know to how to manually solve for the slope of the line, but how do i manually solve for the y-intercept?
I don't know how you got negative values
For the first temperature 0oC, then ΔG = -RT ln Ksp The Ksp you have worked out to be (3.97 * 10-6)2
You should then get ΔG = -8.314 * 273.15 * ln (3.97 * 10-6)2 = 56486.9 J

#### UG

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##### Re: solubility solve for ksp
« Reply #10 on: November 13, 2011, 09:08:38 PM »
I got this...

#### plasticfood

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##### Re: solubility solve for ksp
« Reply #11 on: November 13, 2011, 09:24:41 PM »
opps i forgot the ln...

ok so delta S is -33.543 and delta H is 65649 J? i'm still a bit confused about this b/c in the lab, it says to plot ln ksp vs 1/T, and find these values using the equation of the line. is what you graph basically the same thing? if i were to graph ln ksp vs. 1/T, then my slope would be 775.4, and delta H would be -6446.47 which is not the same.

#### UG

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##### Re: solubility solve for ksp
« Reply #12 on: November 13, 2011, 09:34:00 PM »
Ok I can only assume that if you were to plot ln Ksp vs. 1/T the equation  ΔG = -RT ln Ksp has been rearranged to give ln Ksp = -ΔG/RT. Plotting ln Ksp vs. 1/T would then give you a line with slope -ΔG/R. You would then multiply the slope by R to get G, and then find ΔH and ΔS using the formula ΔG = ΔH - TΔS. I don't know how much of this stuff you have been taught and my assumption could be totally wrong and you are actually meant to be doing something totally different

#### plasticfood

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##### Re: solubility solve for ksp
« Reply #13 on: November 13, 2011, 09:40:13 PM »

yeah i think the formula's rearranged. but it says that the slope is -delta H/R.

i finally got excel to work and plotted ln ksp vs 1/T, and got y = 775.4x - 24.88. from here, i can solve for delta S and H, i think.

#### UG

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##### Re: solubility solve for ksp
« Reply #14 on: November 13, 2011, 09:42:07 PM »
Great, see if they match what I got