![Kb=\frac{[HONH_3^+][OH^-]}{[HONH_2]} =1.1x10^{-8}](/tpl/images/1226/0197/9fc95.png)
![[HONH_2]_0=\frac{1.34g/(33.03g/mol)}{0.500L} =0.0811M](/tpl/images/1226/0197/f8d4a.png)
as the reaction extent which provides the concentration of the hydroxyl ions to subsequently compute the pOH:
![x=[OH^-]=\sqrt{1.1x10^{-8}*0.0811}= 2.99x10^{-5}](/tpl/images/1226/0197/fa871.png)
Any help?
The Kb for hydroxylamine, HONH2, is 1.1 x 10 -8
. What would be the pH of a solutio...
Chemistry, 28.03.2021 04:40, giraffegurl
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?
Answers: 1
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