Question

Any help?

The Kb for hydroxylamine, HONH2, is 1.1 x 10 -8
. What would be the pH of a solution
prepared by placing 1.34 g of HONH2 in 0.500 L of water?

Answer 1

pH = 9.475

Step-by-step explanation:

Hello there!

In this case, according to the basic ionization of the hydroxylamine:

`HONH_2+H_2O\rightarrow HONH_3^++OH^-`

The resulting equilibrium expression would be:

`Kb=([HONH_3^+][OH^-])/([HONH_2]) =1.1x10^(-8)`

Thus, we first need to compute the initial concentration of this base by considering its molar mass (33.03 g/mol):

`[HONH_2]_0=(1.34g/(33.03g/mol))/(0.500L) =0.0811M`

Now, we introduce
`x` as the reaction extent which provides the concentration of the hydroxyl ions to subsequently compute the pOH:

`1.1x10^(-8)=(x^2)/(0.0811-x)`

However, since Kb<<<<1, it is possible to solve for
`x` by easily neglecting it on the bottom to obtain:

`x=[OH^-]=\sqrt{1.1x10^(-8)*0.0811}= 2.99x10^(-5)`

Thus, the pOH is:

`pOH=-log(2.99x10^(-5))=4.525`

And the pH:

`pH=14-4.525\\\\pH=9.475`

Regards!