[bioxoid86]

question is:
if nitrous acid has a Ka at 4.5E-4, and a solution is prepared with 0.5 M of sodium nitrate.
what is the pH.

here is my take on this question so far,

Kb = Kw/Ka = 1E-14/4.5E-4

kb = (x^2)/(0.5-x)

1) does x equals to [H30+] or [OH-]?
2) if kb > ka the solution is basic right?

[bioxoid86]

anyone?

[AWK]

And reaction of hydrolysis (protolysis) is?

[bioxoid86]

that is what i am confused at.
this is what i have so far for the stoichiometry of the equation.(assuming NO3 is -1 charge)
well HNO3 is the conjugate acid of the NaNO3
so

Na + HNO3 + H20 <-> H3O + NO3 + Na

or is it

Na + HNO3 + H20 <-> OH + H2NO3 + Na

i think it is the first one right ??

[AWK]

This is sodium nitrate(III)
NO₂^- + H2O = HNO₂ + OH^-

[bioxoid86]

question is:
if nitrous acid has a Ka at 4.5E-4, and a solution is prepared with 0.5 M of sodium nitrate.
what is the pH.

here is my take on this question so far,

Kb = Kw/Ka = 1E-14/4.5E-4

kb = (x^2)/(0.5-x)

1) does x equals to [H30+] or [OH-]?
2) if kb > ka the solution is basic right?

if i do that kb = 2.22E-11

x=3.33E-6
pOH=5.48
pH=8.52

that can't be yet since ka > kb, the solution is acidic.
but the pH value i calculated 8.52 is >7?

[bioxoid86]

anyone? thanks

[Borek]

that can't be yet since ka > kb, the solution is acidic.

No idea where did you get this misconception from - but it is just misconception. What you have in solution is a weak Bronsted-Lowry base (NO₂^-) so solution must be alkaline.

[agrobert]

This is sodium nitrate(III)
NO₂^- + H2O = HNO₂ + OH^-

Sodium nitrate is NaNO₃

Your equation is for sodium nitrate NaNO₂

Which is not the question.