ASAP 8.0 mL of 0.10 M HC2H3O2 is titrated with 0.10 M NaOH. Perform the following calculations

Question

1. 8.0 mL of 0.10 M HC2H3O2 is titrated with 0.10 M NaOH. Perform the following calculations

-pH at the equivalence point

-Volume of 0.10 M NaOH needed to reach the half-way point

-pH at the half-way point

Answer 1

HC₂H₃O₂ + NaOH ==> NaC₂H₃O₂ + H₂O ... balanced equation for neutralization

moles HC₂H₃O₂ present initially = 8.0 ml x 1 L/1000 ml x 0.1 mol/L = 0.0008 moles

At equivalence, all HC₂H₃O₂ is converted to C₂H₃O₂^- which required 0.0008 moles NaOH.

Volume NaOH needed would be (x L)(0.10 mol/L) = 0.0008 mol and x = 8.0 mls

Total volume at equivalence = 16 mls = 0.016 L

Concentration of C₂H₃O₂^- @ equivalence = 0.0008 mol / 0.016 L = 0.05 M

pH @ equivalence: C₂H₃O₂^- + H2O ==>HC₂H₃O₂ + OH-

Kb = [HC₂H₃O₂][ OH-] / [C₂H₃O₂^- ] = (x)(x)/0.05 ... look up Kb for acetate or Ka for acetic acid. If Ka acetic acid, then find Kb by 1x10^-14/Ka. Use Kb and solve for x which is [OH-] and then find pH

pH at half-way point (i.e. 1/2 equivalence) = pKa for acetic acid (~4.8)