What formula is used to calculate the concentration of base in titration?

The unknown base concentration can be calculated from

Explanation:

An acid–base titration is a quantitative method of analysis for the determination of the concentration of an acid or base by neutralizing it with a standard solution (solution of known concentration) of base or acid.

At the equivalence point, when all unknown  base neutralized by standard acid solution

the following is applied  

No. of equivalent of acid = No. of equivalent of base

Cₐ * Vₐ = Cb * Vb

where,

Cₐ: is the concentration of standard acid solution  (known).

Vₐ: is the volume of acid that neutralizes the base (known).

Cb: is the unknown base concentration (unknown).

Vb: is the volume of base that neutralized with the acid (known).

So, the unknown base concentration can be calculated from

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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