# chemistry

What is the pH of the solution created by combining 12.90 mL of the 0.10 M NaOH with 8.00 mL of the 0.10 M HC2H3O2?

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1. HC2H3O2 + NaOH ==> NaC2H3O2 + H2O

initial:
HC2H3O2 = M x mL = 0.1 x 8.00 mL = 0.8 mmoles.
NaOH = M x mL = 0.1 x 12.90 mL = 1.29 mmoles.
NaC2H3O2 = 0

final:
You have more NaOH than HC2H3O2; therefore, all of the HC2H3O2 will be used and some of the NaOH will remain.
HC2H3O2 = 0
NaC2H3O2 = 0.8 mmoles.
NaOH = 1.29-0.8 = 0.49 mmoles.

concn NaOH = 0.49 millimoles in volume of 12.90 mL + 8.00 mL = 20.90 mL.
concn NaoH = 0.49/20.90 = ??

pOH from that and convert to pH.

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