If the first pKa of H2S is 7.0, what is the pH of a solution containing 10% H2S- and 90% HS-?
To find the pH of the solution, we need to consider the dissociation of hydrogen sulfide (H2S) and its conjugate base, sulfhydryl ion (HS-). The dissociation equation is as follows:
H2S ⇌ H+ + HS-
The pKa value indicates the acidity of an acid and represents the pH at which the acid is 50% dissociated. In this case, the first pKa of H2S is given as 7.0.
Since the solution contains 10% H2S- and 90% HS-, we can assume that the concentration of H2S- is 0.10 and HS- is 0.90 (assuming the total concentration is 1).
Let's assume "x" represents the concentration of H+ ions formed from the dissociation of H2S in the solution.
Using the dissociation equation, we can set up the following equilibrium expression:
Ka = ([H+][HS-]) / [H2S-]
Given that Ka = 10^-7, and [H2S-] is 0.10, and [HS-] is 0.90-x, the equation becomes:
10^-7 = (x)(0.90-x) / 0.10
Now, we can solve this equation to find the value of "x" and determine the concentration of H+ ions. Since x represents the concentration of H+ ions, it can also be used to determine the pH.
By solving the quadratic equation, we find that x = 0.01414
To calculate the pH, we can use the formula: pH = -log[H+]
pH = -log(0.01414)
Using a scientific calculator, the pH value is approximately 1.850.
Therefore, the pH of the solution containing 10% H2S- and 90% HS- is approximately 1.850.