[climbthewall]

I'm really stumped by this question, i was hoping someone might be able to shed a little light on it, it's concerning the debye-huckel limiting law.

The solubility of AgSCN has been studied in aqueous solutions at 298k. The solubility of AgSCN was measured to be 1.069x10^-6 in a solution containing potassium nitrate of unknown concentration. When the concentration of KNO3 was reduced by half the solubility of AgSCN was 1.054x10^-6. Calculate the concentration of potassium nitrate in the origional solution and the solubility product of AgSCN.

i just dont get how to calculate them without knowing the concentration of potassium nitrate, because without it, how do you calculate I, and without that, how do you get the mean activity coefficient. I really would appreciate some help, this has been driving me insane

[Borek]

Without trying to analyze the question in details, obvious line of attack seems to be - write equations describing situation, using I (and I/2) as one of unknowns. Does it yield a system of equations in two unknowns? If yes, solve.

[climbthewall]

Yeah i thought that was what had to be done, but maths is not my strong point and i wasnt sure how to solve simultaneous equations where the answer is an unknown. I wrote this:


1/2 log Ksp = 1.069x10^-6 - 0.509 x I^1/2
1/2 log Ksp = 1.054x10^-6 - 0.509 x I^1/2 / 2

Do  i try and eliminate I and solve for ksp?

[climbthewall]

oh wait, i forgot to rearrange to make it equal to 0

[AWK]

1/2 log Ksp = 1.069x10^-6 - 0.509 x I^1/2
1/2 log Ksp = 1.054x10^-6 - 0.509 x I^1/2 / 2

check these equations

[climbthewall]

is the second I not divided by 2? I thought it might be since I is proportional to concentration here as the charges are all 1?

[climbthewall]

Not meaning to sound pushy but is it at all possible for anyone to give some unambiguous tips. I seem to be going round in circles with this problem.

[climbthewall]

Okay, i did this so far, the ksp i got was close to the standard one i found on the internet...

1/2 logx = log1.069 x10^-6 - 0.509 y
1/2 logx = log1.054 x 10^-6 - 0.509 y/2

1/2 log x = log1.069 x 10^-6 - 0.509 y

log x = log 2.108 x 10^-6 - 1.018 y


1/2log = log1.039x10^-6 - 0.509

1/2 log - log 1.039x10^-6 = -0.509

log x^1/2
    ---------             =-0.509
   1.039x10^-6 

x^1/2
--------              = 10^-0.509
1.039x10^-6 

so x^1/2 = 3.218 x10^-7

ksp = 1.035x10^-13


Is that correct so far?

[climbthewall]

Then,

1/2 log 1.035x10^-13 = log1.069x10^-6 - 0.509 x I^1/2

-6.493 - -5.971
------------------     = I^1/2 = 1.025
-0.509

so I = 1.05

Then I = 1/2 x the sum of all charges squared multiplied by concentrations.

As all of the charges are 1 and the concs are in a 1:1 ratio, then the concentration of each ion = 0.525

So the concentration of kNO3 = 1.05

Is this the correct answer? I would really appreciate if someone could check my working for me?

[Borek]

1/2 logx = log1.069 x10^-6 - 0.509 y
1/2 logx = log1.054 x 10^-6 - 0.509 y/2

Close, but there is a problem. I understand x is Ksp, but I don't understand what y is. From the first equation it seems like a square root of I - but if so, it shouldn't be divided by two in the second equation, as you need which is different from

[climbthewall]

yeah, y = I^1/2

So it should be (I/2) square rooted as opposed to the square root of I / 2, right?

[Borek]

High school algebra:

[climbthewall]

So then multiplying the second equation by 2 will not give I^1/2, so the terms cant be cancelled? My maths really is not great, i dropped it after gcse, How do i cancel the I terms?

[Borek]

You better do something with your math, this is actually very easy problem.

$$ \sqrt {\frac a b} = \frac {\sqrt a}{\sqrt b} /$$

[climbthewall]

So i multiply the entire second equation by square root of 2, this gives me an answer even further away from the standard recorded ksp.