Question

Calculate the pOH of a 15 mL sample of 0.76 mol/L hydrosulfuric acid given that it only dissosiates 92% in solution.

Answer 1

I assume you mean (H^+) = 0.76 x 0.92 = ?
Convert that to pH, then to pOH.

Answer 2

Ms Lena, the 92% dissociation for H₂S (hydrosulfuric acid)* is quiet high for an acid with a Kₐ₁ = 8.9E-8*. (*Ebbing's Gen Chem; 11th Edn.) I calculate the %ionization for a 0.76M H₂S solution to be ~0.034%. Never the less, a 0.76M H₂S solution would have a hydronium ion concentration of 2.6E-4M giving a pH of 3.58. pOH = 14 - pH = 14 - 3.58 = 10.42.

[H⁺]* = SqrRt(Acid Concentration x Ka-value)
*Assuming all of the H⁺ comes from the 1st ionization step as the 2nd ionization step would deliver [H⁺] ~ 1oˉ¹³M range (negligible).
%Izn = [H⁺]/[H₂S] x 100% = (2.6E-4/0.76)100% = 0.034% w/w

Just for consideration, if H₂S were 92% dissociated, then [H⁺] =0.92(0.76M) = 0.6992M => pH = 0.16. This is in the range of very strong acids that ionize 100%. HCl, HBr, HI, HNO₃, HClO₄ & H₂SO₄. Weak acids with Ka-values in the 10ˉ⁵ to 10ˉ¹⁰ range typically have pH values in the 3 to 4 range at ambient conditions.

**hydrosulfuric acid => h ttps://socratic.org/questions/what-is-the-formula-for-hydrosulfuric-acid => H2S