The molarity of sulfuric acid in a fully charged car battery is 5.2 m. when fully discharged the molarity is 4.8m. a 2.00ml sample of battery acid was titrated with 31.26ml of a.621 m sodium hydroxide solution. does the battery need to be recharged

This question requires utilization of  the equivalence point of a reaction, which is a point in the chemical reaction where the two solutions have been mixed in exactly the right proportions. This means that,

, where is the molarity of the acid sulfuric acid, the volume of acid, is the molarity of the base sodium hydroxide and is the volume of the base.

First we write a balanced equation,

.

The stochiometric ratio of is 1:2. Which means 1 mol react with 2 mol. Using the volume and molarity of  given we find the molarity of .

Using moles of we find the molarity

It is fully discharged so the battery needs to be recharged.

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