Acetic acid has a Ka of 1.8 * 10-5. Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations: acetic acid ten times greater than acetate, acetate ten times greater than acetic acid, and acetate=acetic acid.
Match each buffer to the expected pH. pH = 3.74 ; pH = 4.74 ; pH = 5.74
Part B: How many grams of dry NH4Cl need to be added to 2.30 L of a 0.600 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.71? Kb for ammonia is 1.8 * 10-5.
Express your answer with the appropriate units.

A)  acetic acid ten times greater than acetate pH = 3.74

B)  acetate ten times greater than acetic acid pH = 5.74

C) For solution 3: acetate=acetic acid pH = 4.74

Mass NH4Cl = 40.30 grams

Explanation:

Step 1: Data given

Ka is: 1.8 * 10^-5

Solution 1: acetic acid ten times greater than acetate

Solution 2: acetate ten times greater than acetic acid

Solution 3: acetate=acetic acid

Step 2: The pH formula

pH = pKa + log[CH3COO-]/[CH3COOH]

For solution 1:  acetic acid ten times greater than acetate this means ) [[CH3COO-]/[CH3COOH]) has a value of 1/10

pH = pKa + log[CH3COO-]/[CH3COOH]

pH = -log(1.8*10^-5) + log(1/10)

pH = 4.74 -1

pH = 3.74

For solution 2: acetate ten times greater than acetic acid

pH = pKa + log[CH3COO-]/[CH3COOH]

pH = -log(1.8*10^-5) + log(10)

pH = 4.74 + 1

pH = 5.74

For solution 3: acetate=acetic acid

pH = pKa + log[CH3COO-]/[CH3COOH]

pH = -log(1.8*10^-5) + log(1)

pH = 4.74 + 0

pH = 4.74

Part B: Calculate molarity

pOH = pKb + log [H

5.29= -log (1.8*10^-5)  + log [BH+]/[0.600 M]

5.29 = 4.744 + log [BH+]/[0.600 M]

0.546 = [BH+]/0.600

[BH+] = 0.3276 M

Moles NH4+ = 0.3276  M * 2.30 L

Moles NH4+ = 0.75348 moles

Moles NH4Cl = 0.75348 moles

Mass NH4Cl = 0.75348 moles * 53.49 g/mol

Mass NH4Cl = 40.30 grams

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

6.4 cm (max) - 25 cm (min)