Question

To the nearest hundredths, what is the pH at the end of an enzyme-catalyzed reaction if it were carried out in a 0.1 M buffer initially at pH 6.66, and 0.004 M of acid (H+) was produced during the reaction? (The buffer used was a polyprotic acid of the general form: H3A; the three pKas of this polyprotic acid are 2.16, 6.66, and 11.23.)

Answer 1

Solution:

pH matched with pka2 , let acid be H3A hence the buffer would be H2A- and HA2-

We use Henderson equation of buffer

pH = pka2 + log [HA2-]/[H2A-]

6.63 = 6.63 + log [HA2-]/[H2A-]

[[HA2-] =[H2A-] ............(1)

we were mentioned buffer conc is 0.1 M = [H2A-]+[HA2-] .......(2)

by (1) and ( 2)   we get [HA2-]=[H2A-] = 0.05 M

now 0.003 M acid is added , acid H+ reacts with HA2- and we get H2A-

H+ (aq) + HA2- (aq) ---> H2A-(aq)

hence [H2A-] = 0.05+0.004 = 0.054 , [HA2-] = 0.05-0.004 = 0.046 M

now pH = 6.63 + log ( 0.046/0.054)

     = 6.56