Calculate [ H + ] of a 0.330 M solution?

Question

Calculate [ H + ] of a 0.330 M solution of Hydroxylamine HONH2
Kb = 1.10e-8

Answer 1

HONH2 <-> OH-+NH2+
Initial:.33M <-> 0M + 0M
Change:-x<-> +x + +x
__________________________
Equilibrium:.33-x<-> x + x


Kb=concentration of products/concentration or reactants

1.1e-8=x^2/(.33-x)
solve for x: x=6.02e-5, the concentration for OH-

so pOH=-log(6.02e-10)=4.22

Since pH+pOH=14, pH=9.78

[h+]=10^-ph
so 10^-9.78=1.66e-10M [H+]

Answer 2

I am not really sure what Kb stands for but here is what I think.
For every mole of HONH2 you have 3 moles of Hydrogen (you get this by adding the number of hydrogens you see in the formula). So, if the concentraton of your solution is 0.330 M, the concentration of [H+] is 0.330 X 3 = 0.990 M.

Answer 3

The HONH2 will dissociate into OH- and NH2+, of which OH- is the base radical (hydroxyl group). The base dissociation constant given is 1.1*10^-8; this is defined as

[OH-]*[NH2+]/[NH2OH] = 1.1*10^-8,

where the brackets represent molar concentration of the ions. Since one mole of NH2OH will dissociate into equal moles of NH2+ and OH-, let that concentration be [x]; then

[x]^2 / [NH2OH] = 1.1*10^-8; you are given that the compound concentration is 0.330M, so

[x] = √[0.33*1.1*10^-8] = 6.025*10^-5 This then is [OH-]. You are asked for [H+]; but [H+]*[OH-] = 10^-14, the dissociation constant for water, so

[H+] = 10^-14 / 6.025*10^-5 = 1.66*10^-10

The pH is then 9.78

Answer 4

since honh2 is a base we first find out the [OH-]
thus

Kb=1.108
c=0.330M
thus alpha(x)=sqrt kb/c

after finding out x
we find out [OH-]=x2*c

then pOH=log[OH-]


but pH+pOH=14

Answer 5

[H+] = 0.330M
3[H+] = 3*0.330=0.990M

Answer 6

Are you looking for pH or molarity?