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. CHstep 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 Modesto CA escort review 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 you to definitely weak acid dissociates in order to a very brief extent compared to the you to, a good is really so small. formula (1) becomes,
It is not totally dissociated into the a keen aqueous provider thus the second equilibrium is present
Question 13. Define pH. Answer: pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution. pH = – log10 [H3O] (or) pH = – log10 [H + ]
[OH – ] = three times 10 step three Yards. [pH + pOH = 1cuatro] pH = 14 – pOH pH = 14 – ( – journal [OH – ]) = fourteen + journal [OH – ] = fourteen + journal (3 x 10 -3 ) = 14 + record 3 + diary 10 -step 3 = 11 + 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
Question17
pH = – diary [H + ] pH = – record (step one.25 x 10 -2 ) = dos – 0.0969 = step 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(2 x 10 -5 ) = – journal dos – diary (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)