Alright, this time, I know I have not mistyped my answer.
Find the initial concentration of the weak acid or base in an aqueous solution of hydrazine (NH2NH2) with a pH of 10.16. Kb = 1.70 ×10−6 Answer in M
I've tried going from 10^-pH then using that (as my x) in the Kb = x^2/Initial then rearragning that to get x^2/Kb=Initial. I've even tried going from Ka, rearragning everything. But every time either way I get 0.0122899772M
2 answers
Remember the system I use for online homework does not care about significant digits, it goes out to /at least/ 6 decimal places. And I butchered spelling rearranging.
I'll try to remember that but it isn't the usual. So all I can do is see if the set up is ok. Your set up on the earlier problem was ok.
That isn't your x in this hydrolysis equation. Calling NH2NH2 just BNH2, I would convert pH to pOH and use OH as x in the equation of
Kb = (x)(x)/(BNH2)
pH = 10.16; therefore, pOH = 14-10.16 = 3.84 and -3.84 = log(OH^-). I get OH^- = 1.44E-4M for OH^-
.........BNH2 + HOH ==> BNH3^+ + OH^-
I........y...............0........0
C........-x...............x........x
E.......y-x...............x........x
Kb = (x)(x)/(y-x) and solve for y.
That isn't your x in this hydrolysis equation. Calling NH2NH2 just BNH2, I would convert pH to pOH and use OH as x in the equation of
Kb = (x)(x)/(BNH2)
pH = 10.16; therefore, pOH = 14-10.16 = 3.84 and -3.84 = log(OH^-). I get OH^- = 1.44E-4M for OH^-
.........BNH2 + HOH ==> BNH3^+ + OH^-
I........y...............0........0
C........-x...............x........x
E.......y-x...............x........x
Kb = (x)(x)/(y-x) and solve for y.
4 answers