What is the pH of the solution created by combining 2.40 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)?

Question

What is the pH of the solution created by combining 2.40 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)?

I understand how to find the pH for the NaOH + HCl solution.
However, I'm not sure how to find it for HC2H3O2.
I converted all the values into moles:
.00204L x .1 = .00024 mol NaOH
.008L x .1 = .0008 mol HC2H3O2

Then I found the mols of HC2H3O2 left after the reaction, by subtracting .0008-.00024 which gave me .00056 mol left over.

I'm not sure if I find the pH by using the H-H equation where I would use: pH = 4.76 + log[.00056 mol C2H3O2 / .0008 mol HC2H3O2] ..or if I need to do something else to find pH.

Thanks

Answer 1

What is the pH of the solution created by combining 2.40 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)?

Yes. The excess of acetic acid along with the solt (sodium acetate formed) creates a buffered solution. Use the HH equatiion to solve for pH.