Question

1) Calculate the pH in the titration of 50.00 mL of 0.060 M acetic acid (CH3COOH) with a 0.120 M sodium hydroxide, NaOH solution after the addition of the following volumes of base: Ka for acetic acid = 1.8 x 10-5 A) 0 mL pH = B) 10 ml pH =

Answer 1

. CHCOoht 40 - 4 cod + Hot FIJ 0.06 - x 0.06-x [cy, coo] [ų, ot] ka [ch cooth] x? 1-88105 o-of- 1-8X605 (0-08-x) =x? 0.608 x105 – 18x105 (x)=x2 x² + 1.8x105 (x) - 0-108 x 105=0 ax²+bx+cao aba bhayar La - 1.8x105 + / C18110 5) ²461)(-0-108x105) x= - 201) x2 - - 1-8x1osto.0020 785 . - 1.8x 105+0.0020785 Ma E Hy ot72 0.00103025M pHe og 24 ot) = -log (0.00103025) 32.99

© H, cooH: Molarity 20-06M volumea som1 = 0.052 M2 Nach molarity 0.12 M volume = coma o olk no ob molts 0-12 Xool -o-0012 no of moles= 0.06x0.05) et cooth + NaOH CHCOONa tito 0.003 0.001L -0.0012 1-o odl2 to.ool2 0.0018 0.0012 Total volume 0.05+ ordt = 0.062 0.0018 -0.03M sct coolt) a v o.of 0.0012 - - O-OLM {CH COONa] = ka= 1.8 xros pka = -log ka = 10g (1.8x105) = 4.745 [CH, Coona] pH pha tlog Scft coolt] 0.02 24-57 4 4.745 tlog 0.03