## test 2

$\ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}$

Tatum Keichline 2B
Posts: 64
Joined: Fri Sep 28, 2018 12:26 am

### test 2

There was a problem on test 2 asking whether the compound was acidic or basic using K and the other values given. I'm pretty sure we had to use the Van't Hoff equation as well. Does anyone know how to approach this problem?

sonalivij
Posts: 70
Joined: Fri Sep 28, 2018 12:27 am

### Re: test 2

I was confused on this as well. My thinking is that once you find the K2 for the equation you take the square root of it then the -log of it. The answer would be the neutral value. Anything below it is acidic and anything above is basic. This is just a guess though.

Diana Sandoval 1K
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### Re: test 2

How do we know when we have to use that equation?

Felicia1E
Posts: 31
Joined: Fri Sep 28, 2018 12:22 am

### Re: test 2

My TA mentioned that if you use the equation you can determine the new [H+] concentration and then compare it to 10^7. I think if it's higher then it's more acidic, if it's lower its more basic.

Manu Vohra 1L
Posts: 61
Joined: Wed Oct 03, 2018 12:15 am

### Re: test 2

Yes, the Van't Hoff equation would be needed in this case. Essentially you use this equation when you are given a K value for a specific temperature (K1 at T1) and use it to find the K value at another temperature (K2 at T2).

Alexa_Henrie_1I
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Joined: Fri Sep 29, 2017 7:03 am

### Re: test 2

Could someone who got this question correct explain step by step how they did it? I am still confused with this problem.

Gracie Ge 2E
Posts: 28
Joined: Wed Nov 21, 2018 12:19 am

### Re: test 2

you've been given one K value (Kw=10^-14) at 25C, and with the vant hoff equation you can find the other K (we'll call it K2) at 10C.
here's how you do it:
1.plug in the numbers to the vant hoff equation (the temperatures have to be changed to Kelvins):
ln(K2/Kw) = -58kJ/R(1/283K - 1/298K)

2. because we know the value of R (= 8.314JK^-1mol^-1), we can calculate the right hand side of the above equation and get -1.24. (you'll have to change the units of 58kJ to joules though because the units for R is in joules)

which means ln(K2/Kw) = -1.24
we also know Kw = 10^-14
->from that we get
ln(K2/10^-14) = -1.24

3. now we can calculate the value of K2.

regard K2/10^-14 as one entire thing first and you can change
ln(K2/10^-14) = -1.24
to
lnX = -1.24
from that we get
X = 0.289 = K2/10^-14
K2 = 0.289 x 10^-14 = 2.89 x 10^-15

we know that the pH is 7 at 10C
-> we can calculate [H3O+] because pH = -log[H3O+]
-> [H3O+] = 10^-7M
-> [OH-] = K2/[H3O+] = 2.89 x 10^-8 M

4. because there is more H3O+ than OH-, a pH of 7 at 10C is acidic.

Nicole Lee 4E
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Joined: Fri Sep 28, 2018 12:16 am
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### Re: test 2

I found the Kw2 value using the Van't Hoff equation. I square rooted the Kw2 value to find the H3O+ concentration at the new "neutral" pH value. I used -log[H3O+] to find the pH value and compared it with the pH value given in the question. My new pH value ended up being 7.27 and 7.0 is less than 7.27 (considered neutral at the new temperature). Therefore, 7.0 is considered acidic.