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 a weak acid. CH3COOH (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 Hartford escort service, 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 all know one weakened acid dissociates simply to an incredibly short the amount compared to the one to, a good is so quick. equation (1) gets,
It is not completely dissociated in a keen aqueous services so because of this the following harmony is available
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 times ten step three M. [pH + pOH = 1cuatro] pH = 14 – pOH pH = fourteen – ( – diary [OH – ]) = 14 + journal [OH – ] = fourteen + log (three times 10 -step three ) = 14 + diary step 3 + log ten -step three = 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 = – journal [H + ] pH = – record (step 1.twenty-five x ten -dos ) = 2 – 0.0969 = 1.9031
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(dos x 10 -5 ) = – log 2 – record (10 -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)