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

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percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally