A) Calculate [OH^-] in a o.12 M solution of hydrazine, H2NNH2(aq), having K of 3.0x10^-6.
B) what is the percent ionization of the H2NNH2(aq)
I can't get these for the life of mee!
3 answers
If you will show your work I'll look for the error. I think OH^- is about 6E-4 and %ionization = about 0.50%
What i got was :
a) K = (3.0x10^-6)(0.12)/(0.12)(0.12)= 0.6
b) i got 1.90%
a) K = (3.0x10^-6)(0.12)/(0.12)(0.12)= 0.6
b) i got 1.90%
The error is you divided by 0.12 twice; once is all that is needed AND you didn't take the square root.
N2H4 + HOH ==> N2H5^+ + OH^-
Kb = (N2H5^+)(OH^-)/(N2H4)
3.0E-6 = (x)(x)/(0.12)
x^2 = sqrt(3.0E-6*0.12)
I obtained 6.0E-4 for OH^-
Then %ion = [(OH^-)/(0.12)]*100 = ?
I obtained 0.50%
N2H4 + HOH ==> N2H5^+ + OH^-
Kb = (N2H5^+)(OH^-)/(N2H4)
3.0E-6 = (x)(x)/(0.12)
x^2 = sqrt(3.0E-6*0.12)
I obtained 6.0E-4 for OH^-
Then %ion = [(OH^-)/(0.12)]*100 = ?
I obtained 0.50%