posted by anon .
Calculate the pH of the resulting solution if 30.0 mL of 0.300 M HCl(aq) is added to
(a) 35.0 mL of 0.300 M NaOH(aq).
(b) 40.0 mL of 0.350 M NaOH(aq).
I will do one.
first: moles HCL=.030*.3=.009 moles
moles base=.035*.3=.015 moles
so you have excess of .006 moles OH
pOH= -log (.006/.65)
then pH= 14-pOH
Why did you divide by .65?
Wait so, you are getting 0.65 from the total mL but shouldnt this be 65/1000 or 0.065?