Question

If hydrogen sulfide concentrations are greater than 0.15 mg S/L, it is harmful to fish health, calculate the minimum safe pH, if the total sulfide concentration is determined to be 20 micrometers. Assume this is a closed system and there is no H₂S transfer with the atmosphere. pKa₁= 7, pKa₂= 12 for H₂S. Molar mass for sulfur is 32g/mol

Answer 1

Given Terms:

pKa1 = 7, pKa2 =12, Molar Mass of Sulfur = 32g/mol

H2S concentration = 0.15 mg S/L

Total Sulfide concentration = 20 micrometers

Now, let's convert the concentration of H2S

[H2S] (critical) = 0.15 mg S/L x 1 mol/32,000 mg = 4.69 x 10^-8 M

[S] (total) = 20 x 10^-6 M (Given)

The reaction involved is

H2S --> H[+ ion] + HS[- ion]

We know that,

Or

-------> equation 1

As it is clear that, [HS-] = [S] (total) - [H2S] = (2.00 x 10^-5) - 4.69 x 10^-6 = 1.53 x 10^-5 M

The above calculation has been done by assuming certain things,

the concentration of S2- is assumed to be very very less than HS- ([S2-]<<<<[HS-]) and the approximation of some values also taken.

We know that pKa1 = -log(Ka1)

after rearranging, we get,

Ka1 = 10^-pKa1

Ka1 = 10^-7 [because pKa1 = 7 (given)]

Now, putting all the required vales in equation 1, we get,

[H+] = 7 x (4.69x10^-6) / (1.53x10^-5)

[H+] = 3.10 x 10^-8 M

We also know that, pH = -log[H+]

then pH = -log(3.10x10^-8)

pH = 7.51

So, the above value is the minimum safe pH