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

Answer:

pH = 9.475

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:

$Kb=\frac{[HONH_3^+][OH^-]}{[HONH_2]} =1.1x10^{-8}$

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

$[HONH_2]_0=\frac{1.34g/(33.03g/mol)}{0.500L} =0.0811M$

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

$1.1x10^{-8}=\frac{x^2}{0.0811-x}$

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!