posted by alan .
This is my third time to ask this and no one has answered. Please help.
An enzyme-catalyzed reaction is carried out in a 50-mL solution containing 0.1 M TRIS buffer. The pH of the reaction mixture at the start was 8.0. As a result of the reaction, 0.002 mol of H+ were produced. What is the ratio of TRIS base to TRIS acid at the start of the experiment? What is the final pH?
If needed: TRIS(Trizma base) mw= 121.1; pka= 8.3; TRIS-HCl mw= 157.6; pka= 8.3
I kind of have an idea. I start out with the hendersn hasselbalch equation and solve for the ratio ([Tris] / [Tris HCl]). I need help doing this first before i can proceed with the rest of the problem.
Thank you very much.
here is how to solve the problem, put I need help following these instructions. would you show me the math step by step, so I could understand?
Here is how to solve the problem:
Use the Henderson-Hasselbalch equation.
pH = pKa + log ([Tris] / [Tris HCl]). Here the pH is 8 and the pKa of the Tris is 8.3, so you can solve for the ratio ([Tris] / [Tris HCl]).
After the reaction, when 0.002 moles of H+ were produced, this H+ will CONSUME the Tris base to FORM more Tris HCl From the first part of the problem you have the ratio ([Tris]/ [Tris HCl]) = ???. You also know you have 50mL of 0.1M Tris buffer, or 0.005 moles of total Tris (in acid or base forms), which can be written out as "moles Tris + moles TrisHCl = 0.005. Combine this with the equation for the ratio, and solve to find # of moles of both the base and acid form. This will tell you how many moles of acid and base form Tris you had BEFORE the reaction.
Now subtract 0.002 moles from the amount of Tris base, and add 0.002 moles to the amount of TrisHCl you calculated (the H+ consumes base, makes TrisHCl). Now using the same pKa, and the new calculated concentrations of Tris and TrisHCl, you can again use the Henderson Hasselbalch equation, this time solving for pH, the final pH