PH at the equivalence point?

Calculating pH at the equivalence point?

• Hi, Can anyone tell me if my answers are correct? here's the problem: Calculate the pH at the equivalence point for the titration of 20,00mL of benzoic acide by NaOH of 0,0157M knowing that it takes 23,63mL for neutralization. The pKa of benzoic acide is 4,194 at 25°C. I got for the concentration of benzoic acide 0,0185 moles/L Divided by total volume(43,63mL) =0.424M C7H6O2 I then found the Ka of benzoic acid=10^-4,197 which= 6,397*10^-5 10^-14/6.397*10_5= K K=1,56*10^-10=x^2/(0.424-x) so x=8.13*10^-6 so pH=5,09 ? Is this correct??

• Answer:

  moles NaOH = 0.0157 x 0.02363 = 0.000371 = moles benzoic acid C6H5COOH + OH- >> C6H5COO- + H2O total volume = 0.04363 L concentration C6H5COO- = 0.000371 / 0.04363 L = 0.00850 M C6H5COO- + H2O <-----> C6H5COOH + OH- K = Kw /Ka = 1.56 x 10^-10 = x^2 / 0.00850-x x = [OH-] = 1.15 x 10^-6 M pOH = 5.94 pH = 8.06