he concentration of H2S in a saturated aqueous solution at room temperature is approximately 0.1 M. Ka1=8.9x10-8 Ka2=1.0x10-19 Calculate the pH of the solution in 2 sig figs

PH = 4.0

Explanation:

We are given;

Ka1 = 8.9 × 10^(-8)

Ka2 = 1.0 × 10^(-19)

From the 2 values of K given above, we can see that Ka2 is far smaller than Ka1.

Thus, Positive hydrogen ion (H+) will be majorly formed from first dissociation which is Ka1.

Now, the breakdown of the H2S solution is;

H2S⇌[H+] + [HS−]

Thus;

Ka1 = [[H+] × [HS2^(-)]]/(H2S)

HS2^(-) also has a positive hydrogen ion.

Thus, we can rewrite as;

Ka1 = [[H+] × [H+]]/(H2S)

Ka1 = (H+)²/(H2S)

Concentration of H2S is given as 0.1M. Thus;

8.9 × 10^(-8) = (H+)²/0.1

(H+)² = 0.1 × 8.9 × 10^(-8)

(H+) = √(0.1 × 8.9 × 10^(-8))

(H+) = 0.00009433981

Now, PH is gotten from;

PH = -log (H+)

Thus;

PH = -log 0.00009433981

PH ≈ 4.0

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.

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