When an acid solution is titrated with a standardized base solution, separate burets are sometimes used for each solution. which mistake would necessitate emptying and refilling burets, and starting the titration over? a adding distilled water to the titration flask after a solution was measured into it from the buret b overshooting the endpoint c starting with less acid than called for by the procedure d allowing drops of distilled water to stay in the burets while filling them

d) allowing drops of distilled water to stay in the burets while filling them.

Explanation:

Titration is a procedure where we titrated an analyte with a standardized solution and an indicator. For an acid-base titration, knowing the point where the neutralization is completed (pH becomes 7 for a strong acid and base), we can determine the concentration of the analyte.

To do this is necessary to know the concentration and the volume of the standardized solution. So if distilled water is added to the titration flask after a solution was measured, it will not interfere in the study.

By overshooting the endpoint, the titration must be done again, but the volume of the burets only needs to be completed. Same as from the letter "c", the burets will not be emptying and refilling.

But if there is water in the burets, the concentration of the solutions will be changed, it would be more diluted, so it'll be necessary to empty and refill the burets.

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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