What is the pH of the solution created by combining 0.80 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)?

Answer 1

I thought it should be 7? It's not the right answer though.

Answer 2

You have two questions here.
#1. pH of 0.8 mL x 0.1 M NaOH + 8.00 mL of 0.1 M HCl.
moles NaOH = M x L = 0.1 x 0.0008 = 0.00008 moles NaOH.
moles HCl = M x L = 0.1 x 0.008 = 0.0008 moles
NaOH + HCl ==> NaCl + H2O
Now place the moles under the reactants so you can see what is going on. You should note that HCl is in excess; therefore, moles NaCl and H2O formed are 0.00008 and moles HCl left unreacted is 0.0008 - 0.00008 = 0.00072 moles HCl in a total volume of 8.00 + 0.8 mL = 0.0088 L. .
M = moles/L = 0.00072/0.0088 = 0.0818 M
Then pH = -log(H^+) = ??

For the second one, you have HC2H3O2 (a weak acid) and NaOH and the HC2H3O2 is in excess which means that you will have at the end of the reaction a buffered solution consisting of NaC2H3O2 and HC2H3O2. You must use the Henderson-Hasselbalch equation to solve for the pH.
Post your work if you get stuck.

Answer 3

so the henderson eqn is:

pH= pKa +log([conj. base]/[acid])

so do i have to look up pKa for NaOH and then how do i find the conjugate base concentration?