A 0.1 M sodium acetate solution was prepared. The K h = 5.6 × 10 –10
Chemistry · JEE Advanced · NTA Exams — Equilibrium
A 0.1 M sodium acetate solution was prepared. The Kh = 5.6 × 10–10
- The degree of hydrolysis is 7.48 × 10–5
- The [OH–] concentration is 7.48 × 10–3 M
- The [OH–] concentration is 7.48 × 10–6 M
- The pH is approximately 8.88
Answer
(A) The degree of hydrolysis is 7.48 × 10 –5, (C) The [OH – ] concentration is 7.48 × 10 –6 M, (D) The pH is approximately 8.88
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- For the overall reaction H 2 S(aq) ⇌ 2H + (aq) + S 2– (aq), the value of K a is 1 × 10 –22 . The K sp for FeS…
- These 3 questions deal with the following chemical reaction: (green solution) Ni ^ 2+ ( aq ) +6 NH _ 3 ( aq )…
- 100 mL of a solution of X is titrated with a 0.1 M solution of Y giving the following titration curve …
- Which of the following factors will increase solubility of a well known weak base, NH 3(g) , in H 2 O ? NH…
- When two reactants, A and B are mixed to give products C and D , the reaction quotient Q , at the initial…
- Assertion (A) : The pH of the solution at the mid point of the weak acid strong base titration becomes equal…
- When NaNO 3 is heated in a closed vessel, oxygen is liberated and NaNO 2 is left behind. At equilibrium
- During the titration of weak diprotic acid H 2 A against strong base NaOH, the pH of the solution half-way to…