Question 11. Answer: Common ion Effect: When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further. Acetic acid is press the site a weak acid. CHstep step step threeCOOH (aq) \(\rightleftharpoons\) H + (aq)+ CH3COO – (aq)
However, the added salt, sodium acetate, completely dissociates to produce Na + and CH3COO ion. CH3COONa (aq) > Na + (aq) + CH3COO (aq) Hence, the overall concentration ofCH3COO is increased, and the acid dissociation equilibrium is disturbed.
We know from Le chatelier’s principle that when stress is applied to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress. So, in order to maintain the equilibrium, the excess CH3COO – ions combine with H ions to produce much more unionized CH3COOH i.e.,
the equilibrium will shift towards the left. In other words, the dissociation of CH3COOH is suppressed. Thus, the dissociation of a weak acid (CH3COOH) is suppressed in the presence of a salt (CH3COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.
Question 12. Derive an expression for Ostwald’s dilution law. Answer: Ostwald’s dilution law: It relates the dissociation constant of the weak acid (Ka) with its degree of dissociation (?) and the concentration (c). Considering a weak acid, acetic acid. The dissociation of acetic acid can be represented as, CH3COOH \(\rightleftharpoons\) CH3COO – + H + The dissociation constant of acetic acid is,
We realize one to weak acidic dissociates just to an extremely brief the total amount compared to one, a good is really short. picture (1) will get,
This is simply not completely dissociated from inside the a keen aqueous solution so because of this next harmony is obtainable
Question 13. Define pH. Answer: pH of a solution is defined as the negative logarithm of base ten of the molar concentration of the hydronium ions present in the solution. pH = – log10 [H3O] (or) pH = – log10 [H + ]
[OH – ] = 3 x 10 step three Meters. [pH + pOH = 1cuatro] pH = fourteen – pOH pH = fourteen – ( – diary [OH – ]) = fourteen + record [OH – ] = fourteen + record (3 x 10 -step 3 ) = fourteen + record step 3 + log ten -3 = eleven + 0.4771 pH =
Question 15. 50 ml of 0.05 M HNO3 is added to 50 ml of 0.025 M KOH. Calculate the pH of the resultant solution. Solution. Number of moles of HNO3 = 0.05 x 50 x = 2.5 x 10 -3 Number of moles of KOH = 0.025 x 50 x 10 -3 = 1.25 x 10 -3 Number of moles of HNO3 after mixing = 2.5 x 10 -3 – 1.5 x 10 -3 = 1.25 x 10 -3
pH = – log [H + ] pH = – journal (step one.twenty-five x ten -2 ) = 2 – 0.0969 = step one.9031
Question 16. The Ka value for HCN is 10 -9 . What is the pH of 0.4 M HCN solution? Answer: Ka =10 -9 c = O.4M pH = – log [H + ]
? pH = – log(2 x 10 -5 ) = – log dos – journal (ten -5 ) = – 0.3010 + 5 pH = 4.699
Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that. Ka = Kb = 1.8 x 10 -5 Solution.
Question 18. Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak base. Answer: Let us consider the reactions between a strong acid, HCl, and a weak base, NH4OH, to produce a salt, NH4Cl, and water. HCl (aq) + NH4OH (aq) \(\rightleftharpoons\) NH4Cl (aq) + H2O (I) NH4CI(aq) > NH4 + + Cl – (aq)