a)
NH2OH ==> NH2^+ + OH^-
Kb = (NH2^+)(OH^-)/(NH2OH)
Start with 0.50 M NH2OH.
x M ionizes to give x M NH2^+ and x M OH^-. That leaves (NH2OH) = 0.50 M - x.
Plug in the numbers and calculate OH^-
You can solve the quadratic exactly BUT you can assume 0.50 - x = 0.50 (because x is small) and avoid the quadratic equation which will save some grinding of teeth.
Then percent ion = [(OH^-)/0.50]*100
b) is done the same way.
For the second set, both NaOH and LiOH are strong bases. Calculate OH^- from each, add them together, and determine pH and pOH
Same for Ba(OH)2 and RbOH; however, remember that the (OH^-) for Ba(OH)2 is twice the [Ba(OH)2].
Post your work if you want me to check it.
Hello! It's me again.
How do I calculate % ionization for
a.) .50M HONH2 Kb = 1.1e-8
b.) .10M CH3NH2 kB = 4.38e-4
What is [OH-] and pH of these solutions
a.) 0.090M NaOH and 0.090M LiOH
b.) 0.050M Ba(OH)2 and 0.050M RbOH
3 answers
For a.) NH2OH
x^2/.50 = 1.1 x 10^-8 =>
.50(1.1 x 10^-8) = x^2
5.5 x 10^-9 = x^2
5.5 x 10^-9 square root = 7.416 x 10^-5
= x
these are all the steps I have
x^2/.50 = 1.1 x 10^-8 =>
.50(1.1 x 10^-8) = x^2
5.5 x 10^-9 = x^2
5.5 x 10^-9 square root = 7.416 x 10^-5
= x
these are all the steps I have
You are ok as far as you went except I would round the answer off to 7.4 x 10^-5 or carry one more through until the end, then round to 2 figures.
percent ionization = [(OH^-)/0.5]*100
percent ionization = [(OH^-)/0.5]*100
1 answer