Calculate the degree of hydrolysis and pH of 0.02 M ammonium cyanide ( NH 4 CN ) at 298 K. [Ka. of HCN =4.99 × 10-9, Kb for NH 4 OH =1.77 × 10-5 ]

Question

Calculate the degree of hydrolysis and pH of 0.02 M ammonium cyanide ( NH 4 CN ) at 298 K. [Ka. of HCN =4.99 × 10-9, Kb for NH 4 OH =1.77 × 10-5 ] (A) 8.2 (B) 3.2 (C) 9.3 (D) 3.9

Tardigrade Admin · Accepted Answer

Kh=(10-14/4.99 × 10-9 × 1.77 × 10-5)=1.132 It can be seen that hydrolysis constant (Kh) is not small, and for calculating h, the equation used is h =(√Kh/1+Kh)=(√1.132/1+√1.132) =(1.06/1+1.06) =(1.06/2.06)=0.51 Using the formula pH = p Ka- log h+ log (1-h) =- log (4.99 × 10-10)- log (0.51)+ log (1-0.51) =9.3

Q. Calculate the degree of hydrolysis and pH of 0.02M ammonium cyanide (NH4​CN) at 298K. [Ka​ of HCN=4.99×10−9,Kb​ for NH4​OH=1.77×10−5 ]

8.2

3.2

9.3

3.9

Q. Calculate the degree of hydrolysis and $p H$ of $0.02 M$ ammonium cyanide $(N H_{4} CN)$ at $298 K$ . $[K_{a}$ of $H CN = 4.99 \times 1 0^{- 9}, K_{b}$ for $N H_{4} O H = 1.77 \times 1 0^{- 5}$ ]