A solution of methylamine ( CH3NH2 ), a weak base, has an initial concentration of 0.195 M. Methylamine has a Kb of 4.4×10−4. Calculate the pH of the solution.

12

Explanation:

Let's consider the basic reaction of methylamine.

CH₃NH₂ + H₂O ⇄ CH₃NH₃⁺ + OH⁻

The concentration of the weak base is Cb = 0.195 M and the basic dissociation constant is Kb = 4.4 × 10⁻⁴. We can calculate the concentration of OH⁻ using the following expression.

[OH⁻] = √(Kb × Cb)

[OH⁻] = √(4.4 × 10⁻⁴ × 0.195) = 9.3 × 10⁻³ M

The pOH is:

pOH = -log [OH⁻] = -log 9.3 × 10⁻³ = 2.0

The pH is:

pH + pOH = 14

pH = 14 - pOH = 14 - 2.0 = 12

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

answer: mercury

explanation: