CH 3 NH 2 (0.1 mole, K b = 5 × 10 –4 ) is added to 0.08 moles of HCl and the solution is…
Chemistry · JEE Advanced · NTA Exams — Equilibrium
CH3NH2 (0.1 mole, Kb = 5 × 10–4) is added to 0.08 moles of HCl and the solution is diluted to one litre, resulting hydrogen ion concentration is (2005)
- 1.6 × 10–11
- 8 × 10–11
- 5 × 10–5
- 8 × 10–2
Answer
(B) 8 × 10 –11
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- For a sparingly soluble salt A p B q , the relationship of its solubility product ( L s ) with its solubility…
- Assertion (A) : If HA and HB are two weak non-basic acids with K a (HA) b (HB), then, the aqueous solution of…
- An acidic buffer solution can be prepared by mixing the solution of (1981)
- The K sp of Ag 2 CrO 4 is 1.1 × 10 –12 at 298 K. The solubility (in mol/L) of Ag 2 CrO 4 in a 0.1 M AgNO 3…
- Assertion (A) : Increasing the temperature of an aqueous acetic acid solution decreases pH. Reason (R) …
- For the all gas reaction at 1000 K : 2HI ⇌ H 2 + I 2 , K = 0.0344 . If you were to begin with 8.0 M of HI in…
- The solubility product constant of AgCN is 2.5×10 –16 . The acid dissociation constant of HCN is 6.2 × 10 –10…
- 100 mL of a solution of X is titrated with a 0.1 M solution of Y giving the following titration curve …