Question

A 1.0 M H2S solution has a pH = 3.75 at equilibrium. What is the value of Ka?

3.16 × 10−21

3.16 × 10−8

8.89 × 10−5

8.89 × 10−8

Answer 1

To find the value of Ka for a 1.0 M H2S solution with a pH of 3.75, we can calculate the equilibrium concentration of H+ ions using the pH value. Using the equation for Ka, we can then solve for the value of Ka.

Step-by-step explanation:

To find the value of Ka, we need to use the equation for the dissociation of H2S: H2S(aq) → 2H+(aq) + S2-(aq)

Based on the equation, we can see that for every 1 mole of H2S that dissociates, 2 moles of H+ ions are produced. Therefore, the equilibrium concentration of H+ ions is twice the concentration of the H2S solution. Given that the pH of the H2S solution is 3.75, we can calculate the [H+] concentration using the equation pH = -log[H+]. Substituting the pH value into the equation, we get [H+] = 10^-pH = 10^-3.75 = 1.78 x 10^-4 M.

Next, we can calculate the concentration of H2S using the equation for Ka: Ka = [H+][S2-]/[H2S]. Rearranging the equation, we have [H2S] = [H+][S2-]/Ka. Substituting the known values, we get [H2S] = (1.78 x 10^-4)^2 / Ka = 3.17 x 10^-8 / Ka.

Finally, since the given concentration of H2S is 1.0 M, we can set up the equation: 1.0 = 3.17 x 10^-8 / Ka. Solving for Ka, we find that Ka = 3.16 x 10^-8.

Answer 2

When H2S dissociates, the equation is H2S= 2H+ + S2-. In this case, the H+ concentration can be determined from the pH 3.75. taking 10^-pH, H+ is equal to 1.7783 x10^-4 M. Ka is expressed [H+]^2[S2-]/[H2S], this is equal to [1.7783 x10^-4 M]^2 [1.7783 x10^-4 M]/ [1.0-1.7783 x10^-4 M] or 5.62 x10^-12.