Join thousands of students who trust us to help them ace their exams!
Multiple Choice
Determine the pH of 0.115 M Na2S. Hydrosulfuric acid, H2S, possesses Ka1 = 1.0 × 10−7 and Ka2 = 9.1 × 10−8.
A
7.04
B
10.05
C
7.90
D
13.06
0 Comments
Verified step by step guidance
1
Identify that Na2S is a salt derived from the weak acid H2S and a strong base NaOH. In solution, Na2S dissociates completely into 2 Na+ ions and S^2- ions.
Recognize that the S^2- ion is the conjugate base of HS^-, which is the conjugate base of H2S. Therefore, S^2- can undergo hydrolysis in water to form HS^- and OH^- ions, increasing the pH of the solution.
Write the hydrolysis reaction for S^2-: S^2- + H2O ⇌ HS^- + OH^-. Use the equilibrium expression for this reaction: Kb = [HS^-][OH^-] / [S^2-].
Calculate the Kb for S^2- using the relationship between Ka and Kb: Kb = Kw / Ka2, where Kw is the ion-product constant of water (1.0 × 10^-14 at 25°C).
Set up an ICE table (Initial, Change, Equilibrium) to determine the concentration of OH^- produced in the solution. Use the Kb expression to solve for [OH^-], then calculate the pOH and finally the pH using the relationship: pH = 14 - pOH.