a) Kb can be calculated from the relation:
So:
b) We need to build an ICE chart for the equilibrium:
Where the water concentration can be considered as constant:
[NH2OH] | [NH3OH+] | OH- | |
initial | 0.025 M | 0 | 0 |
change | -x | +x | +x |
equilibrium | 0.025 M - x | x | x |
This is a quadratic equation, which can be solved to yield:
The positive result is the one we are looking for, which is equivalent to the concentrations of OH- and of NH3OH-. The concentration of NH2OH is:
c) The pOH is given by:
And the pH:
d) If there were also 0.78 M NH3OH+, the pH could be calculated using the Henderson-Hasselbach equation:
A solution of 0.025 M NH2OH (Ka-1.11 x 10") (20 pt) a) Calculate Kb b) Calculate...
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