Question

QUESTIONS What is the pH of a 0.150 M aqueous solution of ammonium chloride (NH4Cl-)2 Kb (NH3)-1.2 x 10-5 0 A. 4.77 0 B.6.26 U 0 0 O D. 4.95 E. 4.84

Answer 1

For solving this question we will be using the concept of salt hydrolysis. Salt in presence of water( high dielectric constant) undergoes dissociation to form its parent acid and base. We can infer that since it is salt of strong acid(HCL) and weak base(NH4OH) therefore the pH of the salt solution will be less than 7. Detailes solution is given in the image.

to salt of strong acid and weak base NH4Cl + H₂O Ž NH4OH + Her NH4 + cê + H₂O NH4OH + H+ + cé NH + H₂O ² NHLOH + H+ VA - LH ][N 1, b +7 [Nit ] [H₂o Since [H₂o ] is very large, it is taken to be contant and the hydrolysis constant expression is reduced kh= [H +] [NHL OH] [NHA] К = C#tj Lон'] For the dissociation of weak base, NHL OH NH4OH Ť NH + oH The dissociation constant Kb can be written as: Kb = [NH ] [OH- ] - (3) - [NH4OH] Dividing ♡ by ♡ Kw = [Ht] [NH4OH] = kh [NHE and is the h is the degree of hydrolysis molarity of salt of NH 4 I then NH2 + c c-ch H₂O const. - NHL OH + Ht O o ch ch