Question

Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K-1.54 x 10-5), with 0.1000 M NaOH solution after the following additions of titrant. (a) 13.00 mL: pH= (b) 20.50 mL: pH (c) 27.00 mL: pH=

Answer 1

Given data, Kai = 4:65x106 Bazs.1186x1010: PH = 6.97 ule Hnow that [i++] = 10PH = 156.97 =1.07X107, ho (47) (18+)2+(+) (Kaj) + Kas tai 1.07x107 (1.07x177)+(107810) (4:68x1067 4.65X105) 1.07x10? (1-86x10'0). G14x16")* 4.97X16348-64x16 = 1.077107 5:09x1513 PHRA = 2:10X105) » 020-2 Hai[ht? (4+)4 (+) Kas + 70, Kaz 14.65106) C1-07 X1O3) SiOqx1013 - KHA=01976.)

→ Salt o Qu'd + NaOH 2 2.05 o 0.05 COM) - 0.05 2 201 20:50 = 0.05 405 -0.00123 Polt = -109 (04) 2-109 (0.00123) Poh = 2199 We innow that, PH= 14-POH -14-2191 TPH = 11:09)

ci 27 mL added: Millimoles of NaOH = Moenity. Volume - OLX.27 = 2.7 - aud + Nach > salt 22170 0 .017 2 (OH) = i 20127 -0.0212M POH =-109 (OH) .--109 (0.0212) =-(-1.67) POH = 1.67 We know that, PH= 14-POH =14-1167 PH = |233)