The relationship between pH and pOH at 70 ◦C/ 343.15 K

[ChristinaMaria]

Homework Statement

The relationship between pH and pOH, for pure water, at 25 ◦C is
14 = pH + pOH

What is the corresponding relationship between pH and pOH, for pure water, at 70 ◦C?

Homework Equations

Le Châtelier's principle?
[H⁺][OH^-] = 10^-14
[10^-7][10^-7] = 10^-14 (25 ◦C)
[10^-n][10^-m] = 10^-b (70 ◦C)

ΔG◦ = ΔH◦ - TΔS◦ (?)
ΔG = ΔG◦ + RTln(Q) (?)

The Attempt at a Solution

I have tried to use the formula ΔG = ΔG◦ + RTln(Q) for water at 25 ◦C to find the thermodynamic constant Q.
Result:
(at 25 degrees ) ΔG = -48.8 kJ/mol
I then tried to change the equation of ΔG to give Q:
Q = 10^{ΔG / ΔG◦ + RT}

but I realized that I can't do this, as ΔG◦ is dependent on the temperature of the system.

I know that pH decreases at temperature increases due to Le Châtelier's principle. But I don't understand how I can use this to find a corresponding relationship to pH + pOH = ?

Thank you!

 

[Borek]

Are you asked to estimate K_w of water from known thermodynamic data (van 't Hoff's equation comes to mind), or do you just need the value (it is tabulated and not that difficult to find)?

 

[ChristinaMaria]

We are probably supposed to use van'T Hoff's equation! I will try that. Thank you!

 

[ChristinaMaria]

Are you asked to estimate K_w of water from known thermodynamic data (van 't Hoff's equation comes to mind), or do you just need the value (it is tabulated and not that difficult to find)?

Hi again.
I thought I would give an update with my attempt to solve the problem.

I followed your suggestion and used the Van't Hoff equation:

With the following values:
ΔH⊖ = 56000 J/mol, K₁ = 10^-14, R = 8.214 J/mol K, T₂ = 343.15 K and T₁ = 298.15 Kln K₂/10^-14 = 2.99
K₂/10^-14 = 10^2.99 = 977.24
K₂ = 977.24 * 10^-14 = 9.8 * 10^-12

=> (70 C/ 343.14K) [H⁺][OH^-] = 9.8 * 10^-12
pH + pOH = -log( 9.8 * 10^-12 )
[H⁺] = sqrt( 9.8 * 10^-12 ) = 3.13 * 10^-6
pH = -log[H⁺] = 5.5My results:
the relationship between pH and pOH at 70 degrees C is:
[H⁺][OH^-] = 9.8 * 10^-12
and the pH of pure water at 70 degrees C is:
pH = -log[H⁺] = 5.5Unfortunately my results are different from the numerical answers given to us with the exercises, where it says that the pH should be 6.35.
I'm not sure what I did that made it wrong, but I know now that we are supposed to use the Van't Hoff equation as it was confirmed on our class' discussion board.

Thanks again!

 

[Borek]

Hi again.
ln K₂/10^-14 = 2.99
K₂/10^-14 = 10^2.99 = 977.24

What does "ln" mean?

 

[ChristinaMaria]

What does "ln" mean?

I meant "ln" as the natural logarithm to K₂/10^-14. I then raised 10 to the power of K₂/10^-14 to remove the "ln". Is this part wrong?


Update:
I understand what you mean now, I was supposed to raise e to the power of K₂/10^-14, not 10 to K₂/10^-14 like you would do with "log"..!

I used e instead of 10 as the "base", and got the correct answers pH + pOH = 12.7 and pH = 6.35 at 70 degrees C for pure water.
Thank you so much for the help!