Question

Q2 & 4 please??

C6H4COH + HCO CH,OH + HCO, CH,CO2 + H2CO3 CHỌO + H3CO. Ke 4.3x10 2. Sodium carbonate (K) of carbonate ions is 1.8 x 10*) is added to an aqueous s anilinium chloride (Ks of aniline is 4.2 x 10-19. Show that essentially all the anilinium chloride is converted to aniline by calculating the equilibrium constant in water of the reaction CH,NH3* + CO, CHÚNH, + HCO3 3. Fifty millilitres of an aqueous solution contains 3.0 g of substance A. The distribution poefficient of A between ether and water, Ks = c(ether)/(water) = 4:0. Calculate how much of A emains in the water layer after: a) one extraction with 50 mL of ether. -) two extractions with 25 mL of ether. Outline a separation scheme, similar to the one given for this experiment, for separating ture of 2-naphthol (C,H,OH, a very weak acid), 2-naphthylamine (C1H-NH2, a weak bas 1,4-dichlorobenzene. Give all the reaction equations.

only question 2. top 2 equations r not for this question.

Answer 1

2. Ans :-

Acid-base reaction of Na₂CO₃ or CO₃^2- with water is :

CO₃^2- (aq) + H₂O (l) <------------------> HCO₃^- (aq) + OH^- (aq) , K_b = 1.8 x 10^-4 ................(1)

Expression of base dissociation constant (K_b) is :

K_b = [HCO₃^- ].[OH^-] / [CO₃^2-]

Similarly,

Acid-base reaction of C₆H₅NH₃⁺ with water is :

C₆H₅NH₃⁺ (aq) + H₂O (l) <------------------> C₆H₅NH₂ (aq) + H₃O⁺ (aq) , K_a = K_w/K_b = 1.0 x 10^-14 / 4.2 x 10^-10 = 2.38 x 10^-5 ................(2)

Expression of acid dissociation constant (K_b) is :

K_a = [C₆H₅NH₂ ].[H₃O⁺] / [C₆H₅NH₃⁺]

Self ionization water is :

H₂O (l) + H₂O (l) <------------------> H₃O⁺ (aq) + OH^- (aq) , K_w = 1.0 x 10^-14 ................(3)

Expression of ionic product of water (K_w) is :

K_w = [H₃O⁺].[OH^- ]

Adding equations (1) and (2) and then subtract it in equation (3) , we have the required chemical equation is :

CO₃^2- (aq) + C₆H₅NH₃⁺ (aq) <-----------------> HCO₃^- (aq) + C₆H₅NH₂ (aq) , K_c = ?

Expression of equilibrium constant i.e. K_c is :

K_c = [HCO₃^- ].[C₆H₅NH₂] / [CO₃^2-].[C₆H₅NH₃⁺ ]

Now, the relationship between K_c , K_a and K_b is :

K_c = K_a x K_b / K_w

K_c = (2.38 x 10^-5​​​​​​​).(1.8 x 10^-4 ) / (1.0 x 10^-14)

K_c = 4.28 x 10⁵

Hence, equilibrium constant (Kc) of the required reaction is = 4.28 x 105