2. Calculate the pH of a solution prepared by dissolving 0.850 mol of NH3 and 0.300 mol of NH4Cl in water sufficient to yield 2.00 L of solution. The Kb of ammonia is 1.77 x 10-5.

pH = 9.70

Explanation:

It is possible to answer this question using Henderson-Hasselbalch formula:

pH = pka + log [Base] /[Acid]

Base is NH₃ and acid NH₄⁺

Molarity of the compounds is:

NH₃: 0.850mol / 2.00L = 0.425M

NH₄⁺: 0.300mol / 2.00L = 0.150M

To find pka:

Ka×Kb = Kw

Ka = 1x10⁻¹⁴ / 1.77x10⁻⁵ = 5.65x10⁻¹⁰

pKa = -logKa = 9.25

Replacing:

pH = 9.25 + log [0.425] /[0.150]

pH = 9.70

I hope it helps!

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

i think it's c. i'm not sure but i think so.