Question

Titrations 10. Calculate the pH at the equivalence point for the titration above of Sun acid (K=1.8x105) titrated with 0.2M sodium hydroxide. oint for the titration above of 50mL of 0.2M acetic 11. Sketch the titration curve when 50mL of 0.2M acetic acid (K 1.8x105) titrated with 0.2M sodium hydroxide.

Answer 1

10)At equival ence point, ^CH,COOH "oH Рсн соон М сн, соон —DИзьан М юон 50 mL x 0.20 M Hx 0.20 M Volume of NaCH needed to reach equivalence point NaOH 50 mLx 0.20 M 50mL NаОн 0.20 M 1L 0.20 mol No of moles of CH COOH = 50mLx 0.010 mol X- 1000 mL L 1 L 0.20 mol No of moles of NaOH added = 50mLx 0.010 mol X 1000 mL L At equival ence point, pHis determined by the hydrolysis of the conjugate base of the weak acid: No of moles of CH COO no ofmoles of CH COOH 0.010 mol Total volume (50+50) mL 100 mL 0.100 L Concentrati on of CH2COO,CH,COO-7 0.010 mol 0.10 M 0.100 L ICE table CH2COO H20 CH,СООН + ОН- I(M) C(M) E(M) Acidionization constant of CH4COOH, K 1.8x105 0.10 0 0 x +x - X 0.10-x 1.0x104 Ку 1.8x10 Baseionizati on constant of CH2COO,K 5.6x10-10 [сн,соон][он- Baseionizati on constant, Къ CH, COо] (xx) (0.10x) x 7.5x10 M 5.6x10-10 7.5x10 M According to the equilibrium table, OH 1.0x10-1 According to ionic product of water H20*- OH 1.3x10M 7.5x10 -log H0* = -10g (1.3x10) = 8.89 pH