Posted by Mike on Monday, August 10, 2009 at 12:20pm.
You're thinking is on the right track for #1. #2 is the same kind of problem.
For both, calculate moles of HCl initially, (moles = M x L), then calculate moles NaOH added. In the first case, the excess hydroxide will be in moles, divide that by the total volume and that give you molarity of OH^-. Calculate pOH and pH from that.
For #2 it's just the reverse. moles HCl initially - moles OH added leaves an excess of HCl unreacted, moles HCl remaining divided by total volume = molarity and from there you get pH. Post back if you get stuck.
I'm sooo stuck.
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