Sought professional advice. practical 3experiment 1standardization of sodium hydroxide using oxallic acid. useful information. ch04.2h2o m. w. 126.07c=12, h= 1, o= 16, na= 23, ca = 40standardization involves a controlled reaction between a standard solution 300 afixed volume of an analyte which is a solution whose concentration is to bedetermined. it also involves accurate measurements of volumes of solutions ofreactants. apparatusburette, pipette, conical flasks, pipette filler, beakers, burette stands, funnel, clampschemicalssodium hydroxide pellets, oxalic acid dehydrate, phenolphthalein indicatornote. you are provided with sodium hydroxide of unknown concentrationyou are required to prepare 0.1m oxalic acid. page 1 of 2

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