Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.230 M HClO(aq) with 0.230 M KOH(aq). The ionization constant for HClO can be found here.

(a) before addition of any KOH

(b) after addition of 25.0 mL of KOH

(c) after addition of 35.0 mL of KOH

(d) after addition of 50.0 mL of KOH

(e) after addition of 60.0 mL of KOH

Answer 1

), DATE PAGE Helo - coce it f co a ce ke a cox ca a C- ca [ut] a G = cх ke Shox 4+1 = √2.9810-8 x 0.93 +] = 0.667 X10 [+] = 0.8168664 porra - lug (6.816 810) pk = -log(0.816) + 4 = 4:08 Helo & koH coke of Ho somo RimL (0.234 10.239 5 J 11.5 mmoly 5.75in mieles - 5.75 4) 5.75

DATE PAGE This is a buffore Belution H 2 pro & log salt? leida Ha -log (2.9 410-8) + log soff pH = -log 2.9 + 8 2 7.53 To & - Hell somy (0.23 M clook of Co som (0.2314 H₂O 5 mmels 6-9 mmples - 4.6 mmels - 6.9 mmeles he bre & log U Salt? raad 7 : for a boyle 20 9 + lung led to pH = 7.53 + 0.17 = 707

DATE PAGE Holo t somL kok s cooke + 0.23M som lo ke & Ho H₂O 0.23M 11.5mmule 11. simmel 11.5mmal This is a hydrolysis whept. pra I (plew & pro e logo ( 14 + 7.59 bong (1 ks) ] = $(1447-83 + (-0.93)] 2 ( 80-59) = 10:29 e from bomL kot som L requad for - Nardelliclul Eato volume of kolt a fome (leor] > [OX 0.23 = 0.0214 pot 2 - log (0.021) - 1.67 plow a pleeport = 42 14-1.67 = 12.32