Hi Dr. Bob, this is a chemistry question. I don't know how to go about solving this problem:
What is the pH of the solution created by combining 11.40 mL of the 0.10 M NaOH(aq)with 8.00 mL of the 0.10 M HC2H3O2(aq)?
Here's what I have so far:
(0.10M NaOH)(0.0114L)= 0.00114 mol NaOH
0.00114 mol NaOH/(0.0114+0.008)= 0.0588M NaOH
(0.10M HC2H3O2)(0.008L)= 0.0008 mol HC2H3O2
0.0008mol HC2H3O2/(0.0114+0.008)= 0.0412M HC2H3O2
any idea where to go from here? thanks
To make things a little easier to type, let's realize that CH3COOH (acetic acid and written HC2H3O2) can be called HAc where H is the terminal H of the COOH and Ac stands for the remainder of the molecule.
NaOH + HAc ==> NaAc + H2O
moles NaOH you have.
0.1 M x 0.01140 = 0.00114 moles NaOH. (I don't know what your next step means.)
moles HAc = M x L = 0.1 x 0.008 = 0.0008 moles HAc.
So now what happens. The HAc neutralizes 0.0008 moles of the NaOH to leave 0.00114-0.0008 = 0.00034 moles NaOH remaining un-reacted in a volume of
11.4 + 8 = 19.4 mL or 0.0194
So (OH^-) = 0.00034 moles/0.0194 L = ?? and you can get the pOH then pH from this.