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(edited 11 months ago)

Determine the pH of a 100mM aq solution of Tris at 298K

Hi there, not quite sure how to work this one out. I know the balanced equation is:

TRIS (aq) +H2O(l) = TRISH+ (aq) + OH- (aq)

But I'm not sure what to do next

Do you have the value of Kb (the equilibrium constant associated with the reaction you have written an equation for) or Ka given to you?

You may be aware of the relationship Kw = KbKa if you have been given the value of Ka and you may also know the value of Kw at 298 K. From this, you can calculate Kb and use a Kb expression to find [OH^-] and you should be good to go from there.

heres my answer

Original post by thegoldenkimpa

Determine the pH of a 100mM aq solution of Tris at 298K

Hi there, not quite sure how to work this one out. I know the balanced equation is:

TRIS (aq) +H2O(l) = TRISH+ (aq) + OH- (aq)
pka = 8.07 pkw = 14 at 298k

But I'm not sure what to do next

Now that you’ve stated the values of the pKa of TRIS and pKw at 298 K, you may be aware of the following relationships:

pKa = -log[Ka] ==> Ka = 10^-pKa
And similarly, pKw = -log[Kw] ==> Kw = 10^-pKw

Can you now calculate Ka and Kw?

Using the relationship Kw = KbKa, can you now find Kb?

You now want to write an expression for Kb. Since you have written the correct balanced equation for the associated reaction, can you now write an expression for Kb?

Since this is a solution made using just TRIS and water, you can assume that [OH^-] = [TRISH^+]. Can you now rewrite your Kb expression in such a way that it no longer contains a [TRISH^+] in the top of the fraction?

After converting 100 mM to mol dm^-3 by dividing it by 1000, can you now use the value of Kb and your Kb expression to find [OH^-]?

Now using Kw = [OH^-][H^+] and your value of Kw from earlier, you can solve for [H^+] and it should be smooth sailing from here.