Question

What is the percent deprotonation of 1.34 x 10-2 M hydrosulfuric acid H2S (aq) solution? Ka = 9.1 x 10-8

Answer 1

To determine the percent deprotonation of hydrosulfuric acid (H2S) in a given solution, we need to calculate the concentration of the dissociated form (HS-) and compare it to the initial concentration of H2S.

The balanced equation for the dissociation of H2S is: H2S ⇌ H+ + HS-

Let's assume x represents the concentration of HS- formed.

Since the initial concentration of H2S is 1.34 x 10^-2 M, the initial concentration of H+ is also 1.34 x 10^-2 M.

Using the equilibrium expression for the dissociation of H2S, we have: Ka = [H+][HS-] / [H2S]

We can substitute the values into the expression: 9.1 x 10^-8 = (1.34 x 10^-2 - x)(x) / (1.34 x 10^-2)

To solve for x, we can simplify the equation: 9.1 x 10^-8 = (1.34 x 10^-2 - x)(x) / (1.34 x 10^-2) 9.1 x 10^-8 = (1.34 x 10^-2)(x) - x^2 / (1.34 x 10^-2) 9.1 x 10^-8 = 1.34 x 10^-2x - (x^2 / 1.34 x 10^-2) 9.1 x 10^-8 = 1.34 x 10^-2x - x^2 / 1.34 x 10^-2

Rearranging the equation: x^2 - (1.34 x 10^-2)x + (1.34 x 10^-2)(9.1 x 10^-8) = 0

We can solve this quadratic equation to find the value of x using the quadratic formula or other methods. After calculating, we find that x = 2.15 x 10^-5 M.

Now, we can calculate the percent deprotonation of H2S: Percent deprotonation = (concentration of HS- / initial concentration of H2S) * 100 Percent deprotonation = (2.15 x 10^-5 M / 1.34 x 10^-2 M) * 100 Percent deprotonation = 1.60 x 10^-2%

Therefore, the percent deprotonation of the 1.34 x 10^-2 M hydrosulfuric acid (H2S) solution is approximately 1.60 x 10^-2%.