Asked by Lovey dovey
Sodium Azide, (NaN3), the conjugate base of hydrazoic acid, is sometimes added to water to kill bacteria. Given a 0.010 M solution of sodium azide,
A) What is the [H+] concentration from the Azide base rxn?
B) Given a 0.01 M NH4+, what is the [H+]?
C) What is the hydrozoic acid (NH3) half cell potential given 1atm N2(g)?
Please show steps
A) What is the [H+] concentration from the Azide base rxn?
B) Given a 0.01 M NH4+, what is the [H+]?
C) What is the hydrozoic acid (NH3) half cell potential given 1atm N2(g)?
Please show steps
Answers
Answered by
DrBob222
A. One note on this part. NaN3 is NOT the conjugate base of HN3. N3^- is the conjugate base. (NaN3 is the salt of the conjugate base.)
...........N3^- + HOH ==> HN3 + OH-
I........0.01.............0......0
C.........-x..............x......x
E.......0.01-x............x......x
Kb for N3^- = (Kw/Ka for HN3) = (x)(x)/(0.01-x). Solve for x = (OH^-) and convert to H^+.
B. I assume you mean NH4^+ and not ammonium azide. This is done the same way as part A but the hydrolysis is different.
NH4^+ + H2O ==> NH3 + H3O^+
Ka for NH4^+ = (Kw/Kb for NH3) = etc.
C. Can't you find this value in tables?
...........N3^- + HOH ==> HN3 + OH-
I........0.01.............0......0
C.........-x..............x......x
E.......0.01-x............x......x
Kb for N3^- = (Kw/Ka for HN3) = (x)(x)/(0.01-x). Solve for x = (OH^-) and convert to H^+.
B. I assume you mean NH4^+ and not ammonium azide. This is done the same way as part A but the hydrolysis is different.
NH4^+ + H2O ==> NH3 + H3O^+
Ka for NH4^+ = (Kw/Kb for NH3) = etc.
C. Can't you find this value in tables?
Answered by
Yen
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