The following is a list of weak acids and their Ka values. HOCl hypochlorous acid 3.5 x 10^-8H2S hydrogen sulfide 1.1 x 10^-7HCN hydrocyanic acid 4.0 x 10^-10HNO2 nitrous acid 4.5 x 10^-4a. Which acid given above is the strongest? Explain your choiceb. Write the Ka expression for the strongest acid. c. List the conjugate bases of each of the weak acids above and determine their Kb values. d. Which acid has the strongest conjugate base? Explain your choice. e. Determine the pH and pOH of a 0.1M solution of each.

(a) HNO₂ nitrous acid 4.5 x 10^-4 is strong acid

(b) Ka expression

(c) List of Conjugate Bases

OCl⁻, S²⁻, CN⁻, NO₂⁻

(d) cyanide ion is strongest conjugate base.

(e) pH and pOH of Hypochlorous acid = 4.22 & 9.77

     pH and pOH of hydrogen sulfide = 4 & 10

    pH and pOH of  hydrocyanic acid = 5.2 & 8.79

    pH and pOH of  nitrous acid = 2.17 & 11.8

Explanation:

(a) Ka and acidity are directly proportional to each other

hence nitrous acid is the strongest acid since it has highest Ka value.

(b) Ka expression of nitrous acid  

                              HNO₂ → H⁺ + NO₂⁻

                     ⇒      

(c) Conjuagte base for HOCl hypochlorous acid = OCl⁻(hypochlorite ion)

                                 H₂S hydrogen sulfide = S²⁻(sulphide ion)

                                 HCN hydrocyanic acid = CN⁻(cyanide ion)

                                 HNO₂ nitrous acid = NO₂⁻(nitrite ion)

(d) We know according to Bronsted acid base  concept Weak acid have strongest conjugate base

according to this hydrocyanic acid is weak acid so the corresponding conjugate base i.e. cyanide ion is strongest conjugate base.

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; natural resources;