Ionic EquilibriumHard
Question
When 0.1 mole solid NaOH is added in 1lt of 0.1M NH3(aq) then which statement is going to wrong?
(Kb = 2 × 10-5, log 2 = 0.3)
(Kb = 2 × 10-5, log 2 = 0.3)
Options
A.degree of dissociation of NH3 approaches to zero.
B.change in pH would be 1.85
C.conc of [Na+] = 0.1M, [NH3] = 0.1M, [OH-] = 0.2M
D.on addition of OH-, Kb of NH3 does not changes.
Solution
Initial pOH = 
(pKb - log C) = 
(4.7 - log 0.1) = 2.85
Final pOH = 1
Change in pOH = Change in pH = 1.85
(pKb - log C) = (4.7 - log 0.1) = 2.85 Final pOH = 1
Change in pOH = Change in pH = 1.85
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
Which of the following when mixed, will give a solution with pH > 7....Saccharin (Ka = 2 × 10–12) is a weak acid represented by the formula HSac. A 4 × 10–4 mole amount of saccharin is dissol...Consider a weak base ' B ' of ${pK}_{b} = 5.699$. ' x ' mL of 0.02 M HCl and ' y ' mL of 0.02 M weak base ' B ' are mixe...Calculate Kform for HgCl42− if the concentration of Hg2+ is 1.6 × 10–17 M in a solution prepared by dissolving 0.10 mole...Consider a titration of potassium dichromate solution with acidified Mohr′s salt solution using diphenylamine as i...