Question

Rolerences Consider the titration of 100.0 mL of 0.200 M acetic acid K = 1.8 x 10-5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. a. 0.0 mL pH = b. 50.0 mL pH = c. 100.0 mL pH = d. 130.0 mL pH = e. 200.0 mL pH = f. 210.0 mL pH = Submit Answer TEY Another Version 6 ltem attempts remaining.

Answer 1

☺ loomd of 0.200 M acetie Acid Ka = 118 21075 KOH = 0.100m, Plas-dog ( - 4.44 . @ 0.0 ml Koh ~ CH2COOH is weak neid to pH = { pkar & doge = 2.34–20g (0:200) +2.37 +0.349 = 2.719. 5700 1000 so (pas 2.714) | 50.0 ml Xomi total volumes (50+ 100) m Acidr. ito ml 0.200 Mg Cod mol No = 100*0.20º = 0.0200 50 ml 0.100 KOH NO! No - 5048700, 0.0050 Buffer solution - Salt=0.005: eft Aeid. 0.0200 -0.0050) = 0.015. pH = pke + ang Golf) = 4.444 Rog 60.005) 1000 VAcid [pH = 9.96.) @ rooml KOH 100 ml 0.200MCN3COOH a MOL NOS? Tex 100 x0.200, 0.0200 1100 100 ml 0.100 M KOH Mol No= 100x 0.100 -0.010 10000 100 1000 Soulf = 0.0100 Acid = (0.0200 -0.0100) = 0. 0100 pH = pkat log werden = 4.747 Log Coolos) [p" =4.79) 0 130 ml KOH Acid Mol No = 180x0.200 -0.0200 1 heft Acid - (0.0200 -0.013) 20.007 Base Mol No = 130 *0./00 50013 Isafta 0.013. On pH= 4.74 + log 6.007) 1PH 25.00) het end 18.02-0.02) 20 0.02 200X0.100 -0.02 so al no Acid is presen Salt Concenpaction = 0.0281000 Total volume (2004100) = 30ome 7 looo 10.013 @ 200 ml kot = 0.02 .0 or salf= base mol not good resent, orky salf fresmi Acid Mol NO= 0.02 = 0.02 1000 200 - 0.066 sauf pH = 7 th pka thioge DH = 7+2.37 +į hog (06) pH = 8.48 © 210 mL KOH.. 1. Acid Mol No= 0.020 Roase Mol No: 210800 210X0.100 -0.021. so Basic solution form 1000 - Ras fonend Base present = (0.021-0.020) 0.001 310 Base Concenfaefion = 0.001X1000 m Base solution =0.0032 m, por: - cog(0.0032) 2.49 pH=19- 2.49 - |- si o [PP - | F7 )