The ionization constant of ammonium hydroxide is $1.77 \times 10^{- 5}$ at $298 K$ . Hydrolysis constant of ammonium chloride is:

Question

The ionization constant of ammonium hydroxide is 1-77times10-5- 298 K- Hydrolysis constant of ammonium chloride is-5-65times10-12-5-65times10-10-6-50times10-12-5-65times10-13-

Toppr Admin · Accepted Answer

Given that Kb-NH4OH-1-77-xD7-10-x2212-5Kw-10-x2212-14 at 298 K-xA0-Kh-NH4Cl-KwKb-10-x2212-141-77-xD7-10-x2212-5-5-65-xD7-10-x2212-10

The ionization constant of ammonium hydroxide is $1.77 \times 10^{- 5}$ at 298 K. Hydrolysis constant of ammonium chloride is:
$5.65 \times 10^{- 12}$
$5.65 \times 10^{- 10}$
$6.50 \times 10^{- 12}$
$5.65 \times 10^{- 13}$

Given that $K_{b} ({N H}_{4} O H) = 1.77 \times 10^{- 5}$

$K_{w} = 10^{- 14}$ at 298 K

$K_{h} (N H_{4} C l) = \frac{K_{w}}{K_{b}} = \frac{10^{- 14}}{1.77 \times 10^{- 5}} = 5.65 \times 10^{- 10}$

The ionization constant of ammonium hydroxide is 1.77×10−5 at 298 K. Hydrolysis constant of ammonium chloride is:5.65×10−125.65×10−106.50×10−125.65×10−13

Given that Kb(NH4OH)=1.77×10−5Kw=10−14 at 298 K Kh(NH4Cl)=KwKb=10−141.77×10−5=5.65×10−10

The ionization constant of ammonium hydroxide is $1.77 \times 10^{- 5}$ at 298 K. Hydrolysis constant of ammonium chloride is:
$5.65 \times 10^{- 12}$
$5.65 \times 10^{- 10}$
$6.50 \times 10^{- 12}$
$5.65 \times 10^{- 13}$

Given that $K_{b} ({N H}_{4} O H) = 1.77 \times 10^{- 5}$

$K_{w} = 10^{- 14}$ at 298 K

$K_{h} (N H_{4} C l) = \frac{K_{w}}{K_{b}} = \frac{10^{- 14}}{1.77 \times 10^{- 5}} = 5.65 \times 10^{- 10}$