A25.00 ml sample of naoh is titrated with a 0.1000m h2so4 solution. what is the concentration of the naoh if it takes 18.45 ml of the h2so4 solution to reach the equivalence point of the titration?

The chemical reaction involved in this titration process would be:

2NaOH + H2SO4 = Na2SO4 + 2H2O

To determine the concentration of the NaOH solution, we need to first calculate for  the moles of NaOH that is used up in the reaction given the amount of H2SO4 solution that is used. We do as follows:

Moles NaOH = 0.100 M H2SO4 ( .01845 L solution ) ( 2 mol NaOH / 1 mol H2SO4 ) = 3.69x10^-3 mol NaOH reacted

Volume of NaOH = .025 L

Molarity is the number of moles of a solute per liter of solution. We calculate as follows:

Molarity of NaOH = 3.69x10^-3 mol NaOH / .025 L = 0.1476 M

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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