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 2.07, and 0.005 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 2.07, 6.56, and 11.22.) (The enzyme's normal environment is the stomach.)

Answer 1

ans)

from above data that

Buffer maintain the P^H of acids and bases.

P^H = P^Ka of acids.

Here P^H of 2.07 is equal to P^Ka =2.07

We know that formula

P^H = P^Ka+ Log [A^-]/ [HA]

2.07 =2.07 + log [A^-]/ [HA]

log [A^-]/ [HA] =0

Here we get

[A^-]/ [HA] =0

[A^-] = 0, [HA] =0

Here given concentration of phosphate buffer is 0.1M

HA + A^- =0.1M

HA = 0.1M, A^- = 0.1M

0.005M of acid is added to the reaction.It will reacts A^- yields HA.

[A^-] = 0.1-0.005 = 0.095M

[HA] = 0.1 + 0.005 = 0.105M

P^H = P^Ka+ Log [A^-]/ [HA]

= 2.07 + log [0.095]/[0.105]

= 2.07 -0.04346

=2.026

P^H =2.03