The tendency for students is to say that the (OH^-) is 1E-6M but this is only 10 times more than is in H2O when it ionizes; therefore, we must add that in to the mixture. Here is how you do it.
NaOH is a strong electrolyte; therefore it will ionize 100% as follows.
..................NaOH --> Na^+ + OH^-
I.................1E-6..........0...........0
C..............-1E-6.........1E-6.....1E-6
E..................0............1E-6.....1E-6
Then add this to the ionization of water.
...............H2O ==> H^+ + OH^-
I..............liquid.......0........1E-6
C............liquid.......x...........x
E............liquid.......x.........1E-6+x
Kw = (H^+)(OH^-) = 1E-14
(x)(1E-6 + x) = 1E-14
Solve for x = (H^+), evaluate 1E-6+x for OH^- and convert all to pH.
For 2, look on Google and see what you can do for yourself.
1) find the hydronium ion concentration, the hydroxide ion concentration and the pH of a 0.000001 M solution of NaOH.
2) which of salts hydrolyze in aqueous solution : (NH4)2S, Na2So3,(CH3COO)2Ca?
Write the ionic equationss for the hydrolysis reacrions. Predict the pH of solution (pH=7 , pH>7 , ph<7) ?
2 answers
Thank you very much!