Question

What is the pH at the equivalence point if 25.0 ml of 0.015 M benzoic acid (HC₇H₅O₂ , pKa=4.19) is titrated with 0.025 M NaOH?

Answer 1

Answer Benzoic acid is a monoprotic acid. So, Volume of Naot required to reach equivalence point (volume of HC 7 H502) x (molarity of HC7 Hs O2) (molarity of NaOH) (25.0 mL) x (0.015M) (0.025M) 15 mL HC 7 H₂O2 + NaOH NaczHsO2 + H₂O 50g 1 mol of HC7H5O2 Produces 1 mol of ey Hs02 25.0 mL of 0.015 M. HC H4O2 = 0.015*25.0 mol of HC H502 1000 - 0.000375 mol of HG7 H502 TO 50 0.000375 mol of Cats O2 are produced at equivalence point Total volume of solution at equivalence point = (25+15) ML = 40 mL sog molarity of C7 Hs O2 at equivalence point = 0.000375x1000 M = 0.009375 M construct an ICE table corresponding to hydrolysis of C7 H 507 CzHsou + H₂O HC Hs O2 + OH- I(M): 0.009375 O c (M) : x tx tx E(): 0 •00 721s-X [HC7 H502] [OH-] [C7 Hs of 22 - = 1.549x1070 ka 101.19 0.009375-x w, x? + (1-599x1010)x - (1-452x1072) = 0

x -(1549x10")+1(1.549X100)2+(4 X 1X11452x1512) (2x1) (ax + bx+c = 0 -> x = -bi 16-9ac; xxo) za X - 1.205X10M en PH = 14-POH - 14+ log [OH- = 14+ roga = 14 + log (102058 106) = 18.08