## calculating K2?

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

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Katherine Wu 1H
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### calculating K2?

The ionic dissociation of water is given by the following rxn: The ΔH° for the rxn is 58 kJ/mol. The Kw for the rxn at 25 celsius is 10^-14. Is a pH of 7 acidic or basic at 10 celsius?
2H2O(l)⇄H3O^+(aq)+OH^-(aq)

ln K2/K1=-ΔH°/R [1/T2-1/T1]
25 celsius=298.15K
10 celsius=283.15K
58kJ/mol=58,000 J/mol
ln K2/10^-14 = -(58,000 J/mol)/(8.314 J/K.mol)[(1/283.15K)-(1/298.15K)]

pH=-log[H^+]=7
10^-7=0.0000001=[H^+]

K=[OH^-][H^+]
I think I'm supposed to find K2, but I'm not entirely sure how to do that and what to do with it afterwards.

JChen_2I
Posts: 107
Joined: Fri Aug 09, 2019 12:17 am

### Re: calculating K2?

You find K2 by using the Vant Hoff equation you wrote down: ln K2/K1=-ΔH°/R [1/T2-1/T1]
After finding K2, square root it to find the H3O+ concentration. Use the concentration to find the pH. This pH is the neutral pH at 10 degrees C. Compare that number with 7 to see if a pH of 7 is acidic or basic at 10 degrees Celcius.

905373636
Posts: 62
Joined: Thu Aug 01, 2019 12:15 am

### Re: calculating K2?

JChen_2I wrote:You find K2 by using the Vant Hoff equation you wrote down: ln K2/K1=-ΔH°/R [1/T2-1/T1]
After finding K2, square root it to find the H3O+ concentration. Use the concentration to find the pH. This pH is the neutral pH at 10 degrees C. Compare that number with 7 to see if a pH of 7 is acidic or basic at 10 degrees Celcius.

Why do we square root K2 to find [H+]?

JChen_2I
Posts: 107
Joined: Fri Aug 09, 2019 12:17 am

### Re: calculating K2?

905373636 wrote:
JChen_2I wrote:You find K2 by using the Vant Hoff equation you wrote down: ln K2/K1=-ΔH°/R [1/T2-1/T1]
After finding K2, square root it to find the H3O+ concentration. Use the concentration to find the pH. This pH is the neutral pH at 10 degrees C. Compare that number with 7 to see if a pH of 7 is acidic or basic at 10 degrees Celcius.

Why do we square root K2 to find [H+]?

Because the [H+] and [OH-] concentrations will be equal at equilibrium.

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