[erv10s]

Question:

If 2 x 10^(-6) moles of R-O-SO₃^-Na⁺ are added to 1 mL of 50 mM Tris-HCl (pH 8.1) and completely hydrolyzed to ROH = SO₄^2- + H⁺ by the addition of a sulfatase enzyme, what will be the final pH of the solution?


I am having a hard time just starting this problem. From previous help that I got, I know that I think need to find the number of moles of H+ from the enzyme and then use that information somehow to find the final pH using the numbers for Tris-HCl in the Henderson-Hasselbalch equation. Basically though, I need some guidance in starting this problem.

Thanks again!

[Yggdrasil]

What you have posted is the correct approach to solving the problem, although it's missing some steps.  First, I would say, calculate the initial amounts of tris base and tris acid using the Henderson-Hasselbalch equation.  Next account for the reaction of tris base with the acid, then use the new numbers of [acid] and [base] to recalculate the pH.

[Borek]

Just remember: like in the other question, assume that all protons from hydrolysis are consumed by the pronation reaction.

[erv10s]

Here is my work so far:

Since pH = pKa (8.1) for Tris HCl, the initial amounts for Tris base and Tris acid would be 0.05 M.

Then,

Tris base (5 x 10^-5 mol) + R-O-SO₃^-Na⁺ (2 x 10^-6 mol) ---> ROH + SO₄^2- + H⁺

Which leaves 4.8 x 10^-5 mol of Tris base and 2 x 10^-6 mole of the conjugate.

pH = 8.1 + log (2 x 10^-6/4.8 x 10^-5)
pH = 6.7

[Borek]

Since pH = pKa (8.1) for Tris HCl, the initial amounts for Tris base and Tris acid would be 0.05 M.

No, 50 mM buffer means that sum of their concentrations is 0.05.

[erv10s]

Oh ok, so

8.1 = 8.1 + log (x/.05-x)

So the concentration for the Tris base and Tris acid would both be .025 M.

Then is it correct if I proceed like before?

Tris base (2.5 x 10^-5 mol) + R-O-SO^3-Na⁺ (2 x 10^-6 mol) ---> ROH + SO₄^2- + H⁺

Which leaves 2.3 x 10^-5 mol of Tris base and 2 x 10^-6 mole of the conjugate.

pH = 8.1 + log (2 x 10^-6/2.3 x 10^-5)
PH = 7.03

[Borek]

Oh ok, so

8.1 = 8.1 + log (x/.05-x)

So the concentration for the Tris base and Tris acid would both be .025 M

Correct.

Now, you have added 2x10^-6 mole of strong acid. How does the concentration of TRIS base change? How does the concentration of TRIS acid change?

[erv10s]

The concentration of Tris base decreases by 0.002 M, and the concentration of Tris acid increases by 0.002 M.

So,

pH = 8.1 + log (.027/.023)
pH = 8.17

[Borek]

pH = 8.1 + log (.027/.023)
pH = 8.17

Just by looking: you have added strong acid to the pH 8.1 solution, and pH went up? I don't buy it.

[erv10s]

Yes, my mistake...clearly that does not make any sense. Well, am I at least right in thinking that the base is decreasing by .002 M and the acid increases by .002 M ?

The only way I'm getting a decrease in pH is if I switched to log (.023/.027) or did it like my previous incorrect calculation : pH = 8.1 + log (.002/.023).

[Yggdrasil]

Given that the HH equation says pH = pKa + log ([base]/[acid]), you should use log (.023/.027).

[erv10s]

Ah..thanks I did not even see that mistake.

[Borek]

Note: it was more or less obvious what is wrong, my main point was that it is quite often enough to look critically at the result to see that it must be incorrect.