what would the ph be at the end of an enzyme-catalyzed reaction if it were carried out in a 0.1 m phosphate buffer of pka 6.86 and 0.005 m of acid was produced during the reaction

Ph of the enzyme -catalyzed reaction = 6.77

Explanation:

Given data:

concentration phosphate buffer =  0.1 m

pka. = 6.86  

concentration of acid in the reaction = 0.005 m

Determine the  Ph at the end of the reaction

For a balanced buffer solution ; Acid amount = base amount

first we will determine the concentration of acid and base in relation to the buffer

acid = 0.1 / 2 = 0.05

base = 0.1 /2 = 0.05

The chemical reaction of the buffer can be expressed as attached below

PH = pka + log =

     = 6.86 + log ( 0.8181 ) = 6.77

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

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