1.

The ionization constant of ammonium hydroxide is 1.77xx 10^(-5) at 298 K. Hydrolysis constant of ammonium chloride is

Answer»

`5.65 XX 10^(-10)`
`6.50 xx 10^(-12)`
`5.65 xx 10^(-13)`
`5.65 xx 10^(-12)`

Solution :`K_(h) = (K_(w))/(K_(b)) = (1 xx 10^(-14))/(1.77 xx 10^(-5)) = 5.65 xx 10^(-10)`


Discussion

No Comment Found

Related InterviewSolutions