Calculate the pH of the resulting solution if 30.0 mL of 0.300 M HCl(aq) is added to

Question

(a) 35.0 mL of 0.300 M NaOH(aq).
(b) 40.0 mL of 0.350 M NaOH(aq).

Answer 1

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

Answer 2

Why did you divide by .65?

Answer 3

Wait so, you are getting 0.65 from the total mL but shouldnt this be 65/1000 or 0.065?