Question

5. What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of an aqueous weak acid requires 29.80 mL of 0.0567 M NaOH? Ka = 3.2 x 10-4 for the weak acid.

Answer 1

At equivalence point, wx = NACH 25 mL x Myx = 29.80 mL < 0.0567 M Concentration of acid needed to reach equivalence point, 29.80 mL < 0.0567 M Myx= s 0.0676 M 25 mL No of moles of HX = 25 mLx_1L 0.0676 mol ML * 1000 mL ** L 101 = 0.00167 mol . 1L 0.0567 mol No of moles of NaOH added = 29.80 mL = 0.00167 mol 1000 mL L At equivalence point, pHis determined by the hydrolysis of the conjugate base of the weak acid: No of moles of X-=no of moles of HX = 0.00167 mol Total volume = (25+29.8) mL = 54.80 mL = 0.0548 L Concentration of X, [x] = 0.0548 L ICE table x + H2O → HX + OH- I(M): 0.0305 C(M): -X to to E (M): 0.0305- X Acidionization constant of HX, K, = 3.2x10+ 1.00x10-14 Baseionization constant of X,K 0.00167 mol = 0.0305 M + 3.2x10+ = 3.1x10-11 Baseionization constant, [HX]OH-] Ko = [x] 3.1x10-12 = (x) (x) (0.0305- x) x=9.76x10-'M According to the equilibrium table, OH 1 = (x) M = 9.76x10-'M K 1.00 x10-14 According to ionic product of water, H0 OH) 9.76x10-7 = 1.02x10-1 pH = -log[1,01] =-log(1.02x10 ") = 7.99 x 8.0