The weak base OCl^- (aq), hypochlorite, will react with strong acid, H_3O^+ (aq)
ID: 523059 • Letter: T
Question
The weak base OCl^- (aq), hypochlorite, will react with strong acid, H_3O^+ (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 K_c to each). Add the two equilibria to yield the net reaction; now, find the value of K_c for the net reaction. Please propose whether or not the net reaction will proceed 100% from (1 rightarrow r); please explain your proposed answer. 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 K_a for HOCl can be found in Table 14.2.Explanation / Answer
5) OCl- is the weak acid of hypochlorous acid (HOCl) and forms the following eequilibrium with water as below:
OCl- (aq) + H2O (l) -------> HOCl (aq) + OH- (aq)
Kc1 = [HOCl][OH-]/[OCl-][H2O] …..(1)
Hydronium ion (H3O+) combines with hydroxide (OH-) to form water as shown below:
H3O+ (aq) + OH- (aq) ------> 2 H2O (l)
Kc2 = [H2O]2/[H3O+][OH-] …..(2)
Combine the two equations above to yield the equilibrium constant for the neutralization reaction as
OCl- (aq) + H3O+ (aq) <=====> HOCl (aq) + H2O (l)
The equilibrium constant for the above reaction is given by
Kc’ = Kc1*Kc2 = [HOCl][H2O]/[OCl-][H3O+]
In an aqueous solution, [H2O] is large (approx., 55.5 M) and hence the concentration of [H2O] is virtually assumed to be constant and we define the modified equilibrium constant Kb for the above reaction as
Kb = Kc’/[H2O] = [HOCl]/[OCl-][H3O+] …..(3)
The equilibrium reaction will proceed 100% in the direction of the product (i.e, from left to right) because a strong acid (H3O+) combines with a base to give a weak acid (HOCl).
6) The net equation taking place is
OCl- (aq) + H3O+ (aq) ------> HOCl (aq) + H2O (l)
As per the balanced stoichiometric reaction above,
1 mole OCl- = 1 mole H3O+ = 1 mole HOCl.
Identify that aqueous HCl is the source of H3O+. Therefore, 20 mmole of H3O+ will completely neutralize 20 mmole of OCl- to form 20 mmole of HOCl.
The volume of the solution is 50.00 mL.
Therefore, concentration of the HOCl formed = (20 mmole)/(50.00 mL) = 0.4 mol/L = 0.4 M.
HOCl is the weak acid of OCl- and establishes equilibrium with water. We set up the equation and the ICE chart as below:
HOCl (aq) + H2O (l) <====> H3O+ (aq) + OCl- (aq)
initial 0.4 0 0
change -x +x +x
equilibrium (0.4 – x) x x
The equilibrium constant for the above reaction is the acid-dissociation constant for HOCl and is written as
Ka = [H3O+][OCl-]/[HOCl] = (x).(x)/(0.4 – x)
Find the value of Ka from table (I used internet sources) to be 3.5*10-8; therefore,
3.5*10-8 = x2/(0.4 – x)
Use the small x approximation. Since Ka is small (of the order of 10-8), we can assume that x << 0.4 and hence (0.4 – x) 0.4. Plug in values and write
3.5*10-8 = x2/0.4
===> x2 = 1.4*10-8
===> x = 1.183*10-4
Therefore, [H3O+] = 1.183*10-4 mol/L and pH = -log [H3O+] = -log (1.183*10-4) = 3.927 3.93.
The pH of the solution after the neutralization is 3.93 (ans).
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