a. Please show how the weak base OCl - (aq), hypochlorite, will react with stron

Question

a. Please show how the weak base OCl- (aq), hypochlorite, will react with strong acid, H3O+ (aq), in a neutralization reaction. To answer this, please create two equilibria (one for the weak base dissociation, the other for the reaction of hydronium ion with dissociated hydroxide to form water, please assign values of Kc to each). Add the two equilibria to yield the net reaction; now, find the value of Kc to each). Add the two equilibria to yield the net reaction; now, find the value of Kc for the net reaction. Please propose whether or not the net reaction will proceed 100% from (L ----> R); please explain your proposed answer.

b. Using your net reaction from above, please determine the pH of the solution formed when 20.00 mmol of OCl- is neutralized by 20.00 mmol of HCl (aq) in a solution of volume 50.00 mL. Please rewrite your net equation and create an ICE chart to determine the mmol of products. The value of Ka for HOCl can be found in Table 14.2.

********************IF YOU PREVIOUSLY ANSWERED THIS QUESTION IN ANOTHER POST DO NOT ANSWER THIS POST PLEASE************************

Explanation / Answer

weak acid:

HClO(aq) + H2O(l) <--> H3O+(aq) + OCl-(aq) K = 3.5*10^-8

the water equilbirium

H2O(l) + H2O(l) <--> H3O+(aq) + OH-(aq) K = 10^-14

then, the neutralizaitno is given as:

OCl-(aq) + H3O+(aq) <--> HOCl-(aq) + H2O(l)

for thi, kb

Kb = Kw/Ka = (10^-14)/(3.5*10^-8) = 2.85714*10^-7

The reaction will not go 100%

it actually remains fractional as

OCl- and HOCl

b)

mmol of OCl- + mmol of HCl = 20 mmol of HOCl produced

HOCl hydrolysises:

HOCl <--< H+ + OCl-

Ka = [H+][OCl-]/[HOCl]

ICE:

initially [H+] = 0, [OC-] = 0, [HOCl] = 20/50 = 0.4 M

Change [H+] = +x, [OC-] = +x, [HOCl] = - x

Equilibrium [H+] = 0+ x , [OC-] = 0+ x, [HOCl] = 0.4- x

substitute

3.5*10^-8 = x*x/(20/50-x)

x = 1.18*10^-4

[H+] = 1.18*10^-4

pH = -log(1.18*10^-4) = 3.9281

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.