Calculating pH from deltaG° and Ion Concentrations

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The discussion centers on calculating the pH required for the equilibrium of the reaction I2(s) --> I- (aq) + IO3- (aq) at 298 K, given the standard Gibbs free energy change (deltaG°) of -153.8 kJ/mol and ion concentrations of [I-]= 0.5 M and [IO3-]= 0.5 M. The equilibrium constant (Keq) was determined to be 9.113 x 10^26. The correct pH calculation, derived from the concentration of hydroxide ions ([OH-]), yields a final pH of 9.21. The method involved solving for [OH-] using the equilibrium expression and converting it to pH using the formula pH = 14 + log[OH-].

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RNix25
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I've been working on this one for a few days, and I just can't seem to get the right answer.

Consider the reaction I2(s) --> I- (aq) + IO3- (aq)
Balanced= 3I2 + 6OH- --> 5I- + IO3- + 3H2O

deltaG°= -153.8 kJ/mol

The question: What pH is required for the reaction to be at equilibrium at 298 K when [I-]= 0.5 M and [IO3-]= 0.5 M

I know the answer is 9.21, but I just can't figure out how to get there. I started by finding Keq, which is 9.113 * 10^26, but I'm stumped from there. I would be extremely appreciative for a walkthrough to the final answer.
 
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Sounds like this should go in the homework help section.

Your K value looks correct. Write out K in terms of the concentrations of species raised to the power of their stoichiometric coefficients and solve for [OH-]. pH = 14 + log[OH-], I get 9.206 using your values.
 
I slept on it and when I woke up, I looked over it and figured out where I was going wrong. Had one of my concentrations wrong, so I kept getting the wrong value when I worked it. I already had the answer, I just couldn't figure out why my math wasn't producing it. Anyway, your method was exactly right, so thanks for the help.

Don't know protocol on these boards, but if it is proper, a mod can close/delete this thread.
 

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