Question

Hydrolysis constant (Kh) of 0.1
ammonium cynide solution at 250
C when ionization constant of
HCN (ka) = 7.2 x 10-10, ionization
constant of ammonium hydroxide
(Kb ) = 1.77 x 10-5 and ionic
product of water (Kw) at 250 C =
1.0 x 10-14 will be..
PRN 1172017801 NOTE:​

Answer 1

Explanation:

8.0×10-10

s

d

d

d

d

d

d

t

t

y

y

y

y

Answer 2

Answer:

The ammonium cyanide solution's hydrolysis constant, $K_{h}$, measured is $0.885$.

Explanation:

Given,

The ionization constant of $HCN$, $K_{a}$ = $7.2\times 10^{-10}$

The ionization constant of $NH_{4}OH$, $K_{b}$ = $1.77\times 10^{-5}$

The ionic product of water, $K_{w}$ = $1\times 10^{-14}$

The ammonium cyanide solution's hydrolysis constant, $K_{h}$ =?

As we know,

• The hydrolysis constant, $K_{h}$ = $\sqrt{\frac{K_{w} }{K_{a}\times K_{b} } }$

Therefore,

• $K_{h}$ = $\sqrt{\frac{K_{w} }{K_{a}\times K_{b} } }$
• $K_{h}$ = $\sqrt{\frac{1\times 10^{-14} }{7.2\times 10^{-10} \times 1.77\times 10^{-5} } }$
• $K_{h}$ = $\sqrt{\frac{ 10^{-14} }{12.744\times 10^{-15} } }$
• $K_{h}$ = $\sqrt{\frac{ 10 }{12.744 } }$
• $K_{h}$ = $\sqrt{0.784}$ = $0.885$

Hence, the hydrolysis constant of ammonium cyanide, $K_{h}$ = $0.885$.