Question

The hydronium ion concentration of an aqueous solution of 0.489 M hydroxylamine (a weak base with the formula NH2OH) is ...

H3O+= _________M

Answer 1

Given [NH2OH]=0.489 M.

Kb=9.1 x10^-9

Now ICE table,

NH2OH + H2O <---------> NH3OH+OH^-

Initial. 0.489 . 0 . 0

Change. - x . +x. +x

Equilibrium. 0.489-x . x. x

Kb=[NH3OH][OH^-]/[NH2OH]

9.1x10^-9=x^2/(0.489-x)

Since Kb<<<x, then 0.489-x~ 0.489.

Therefore x^2=0.489 x 9.1x10^-9

x=(4.4499x10^-9)^(1/2)

x=6.67x10^-5.

Therefore [OH^-]=x=6.67x10^-5 M

Since [H3O^+][OH^-]=1x10^-14

[H3O^+]=(1x10^-14)/(6.67x10^-5)=1.499x10^-10

[H3O^+]=1.499x10^-10 M.

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