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 6.73, 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 2.17, 6.73, and 11.27.)

Answer 1

Solution:

The computation are as expressed below,

From the given data pH of 6.73 is similar to given pKa value of 6.73 hence the making of buffer solution is possible,

pH = pKa + log [A^-] / [HA]

6.73 = 6.73 + log [A^-] / [HA]

1 = [A^-] / [HA]

[A^-] = [HA] ----> eq.(1)

for given 0.1M buffer

[A^-] + [HA] = 0.1

plugging eq.(1) we get,

[HA] + [HA] = 0.1

2 [HA] = 0.1

[HA] = 0.05 M & [A^-] = 0.05 M

as given 0.004 M acid formed so we get,

[A^-] = 0.05 - 0.004 = 0.046 M

[HA] = 0.505 - 0.004 = 0.046 M

now plugging values in the below equation we get pH,

pH = pKa + log [A^-] / [HA]

pH = 6.73 + log 0.046 / 0.046

pH = 6.73