Question

When an organic sulfate ester is hydrolyzed by a sulfatase enzyme, a hydrogen ion is produced;

R-O-SO₃^- + H₂O ----> ROH + SO₄^-2 + H⁺

The above reaction was carried out in a 1.0 mL of a .02 M Tris buffer, pH 8.10, containing 0.01 M ester. At the end of 10 minutes, the pH of the reaction mizture decreases to 7.95. How many micromoles of ester were hydrolyzed during the 10 minute reaction period? pKa of Tris = 8.10.

Answer 1

Using Henderson Hasselbach equation:

pH = pKa + log([salt]/[acid])

Initially, pH = pKa, which means that [salt] = [acid]

Also, it is given that buffer conc. is 0.02 M, which means:

[salt] + [acid] = 0.02

So, initially,

[salt] = [acid] = 0.01

When ester is hydrolyzed, protons are produces.

Assume that 'x' moles of protons are produced, this means moles of acid increase and that of salt decrease by this much amount. So we have:

7.95 = 8.10 + log( (0.01-(x*1000))/(0.01+(x*1000)) )

Solving we get:

x = 0.00000171 = 1.71 micromoles

So, these many moles of protons were produced by the reaction.

Since 1 mole of proton is produced by the hydrolysis of 1 mole of ester, so number of moles of ester that were hydrolyzed = 1.71 micromoles

Hope this helps !