Posted by Mandy on Monday, September 14, 2009 at 10:12am.
This has a very good explaination. Go thru his examples, especially at the end.
On c, find the OH concentration, then the pOH, then you know pH+pOH=14 and you can solve for pH from that.
a) 0.35M HCl (monoprotic strong acid) produces a 0.35M [H+] solution.
pH = -log(0.35)
c) 0.35M NaOH (strong base) acid produces a 0.35M [OH-] solution.
pOH = -log(0.35)
pH = 14-pOH
b) 0.35M HC2H3O2 (WEAK ACID) does not dissociate completely. Its Ka = 1.8x10^-5
[H+] = sqrt[(Ka)(0.35)]
pH = -log[H+]
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