In another experiment, the student titrated 50.0 mL of 0.100 M HC, H,O, with 0.100 M NaOH(aq). Calculate the pH of the solution at the equivalence point

Eqv Pt pH = 8.73

Explanation:

    HOAc                   +            NaOH            =>            NaOAc              + H₂O

50ml(0.10M HOAc)  +  50ml(0.10M NaOH) => 100ml(0.05M NaOAc) + H₂O

For neutralized system, 100ml of 0.05M NaOAc remains

NaOAc => Na⁺ + OAc⁻

Na⁺ + H₂O => No Rxn

          OAc⁻  +  H₂O  => HOAc + OH⁻

C(i)   0.05M              0M      0M

ΔC        -x                   +x        +x

C(f)    0.05-x      

       ≅ 0.05M              x          x

Kb = Kw/Ka = [HOAc][OH⁻]/[OAc⁻] = 1 X 10⁻¹⁴/1.7 X 10⁻⁵ = (x)(x)/(0.05M)

=> x = [OH⁻] = SqrRt(0.05 x 10⁻¹⁴/1.7 x 10⁻⁵) = 5.42 x 10⁻⁶M

=> pOH = -log[OH⁻] = -log(5.42 x 10⁻⁶) = 5.27

pH + pOH = 14 => pH = 14 - pOH = 14 - 5.27 = 8.73 Eqv Pt pH

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there are two significant figures.