Airlia Veteran 185 Posts user info edit post |
I feel for you on this one. The lab's moving faster than lecture for most Ch201 sections.
You have 10 ml x 0.1 M HCl for 1 mmol H3O^1+. You have 5 ml x 0.1 M NaOH for 0.5 mmol OH^1-.
Set up a reaction table for the reaction, assume 100% reaction and determine limiting reagent: H3O^1+ + OH^1- --> 2H2O 1.0 mmol 0.5 mmol 0 mmol initial -0.5 mmol -0.5 mmol +0.5 mmol delta --------------------------------------- 0.5 mmol 0 mmol 0.5 mmol
So you've got 0.5 mmol H3O^1+ in 10 + 5 = 15 mL solution. This would make your [H3O^1+] = 0.5 mmol/15 mL = 0.033333 M
pH = -log [H3O^1+]. I'll leave the last step to you.
I thought the lab manual might describe this pretty well, but I'm at home without the manual so I'll have to check tomorrow. 2/27/2006 8:34:33 PM |