Aweak acid (ha) has a pka of 4.001. if a solution of this acid has a ph of 4.110, what percentage of the acid is not ionized? (assume all h in solution came from the ionization of ha.)

99.9224 % of the acid is not ionized.

HA + H₂O ⇌ A⁻ + H₃O⁺

I: ___c 0 0  

C: _-αc +αc __+αc

E: c(1-α) αc ___αc

pH = 4.110

[H₃O⁺] = αc = 10^(-4.110) mol/L = 7.76 × 10⁻⁵ mol/L

α = 7.76 × 10⁻⁵

1 – α = 1 - 7.76 × 10⁻⁵ = 0.999 224 = 99.9224 %

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.