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

indium, and thallium have three electrons i think

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."