For all of
the following questions 10.00mL of 0.187 M acetic acid
(CH_{3}COOH) is titrated with 0.100M KOH (The K_{a} of acetic acid is 1.80 x 10^{5}) Region 1: Initial pH Before any titrant is added to the starting material 

Tabulate the concentration of the species involved in the equilibrium reaction, letting x = [H^{+}] at equilibrium (Do not calculate “x” yet)
Use the K_{a} expression and determine the concentration of H^{+} and the pH of this solution.


Region 2:
Before the Equivalence Point 12.83mL of the 0.100M KOH has been added to the starting material 

First, fill out the BCA table to determine the moles of material at this point in the titration. (Fill in actual values, do not use x. Also be sure to use moles)
The resulting concentrations of CH_{3}COOH and CH_{3}COO^{} after the reaction here are:
Use these calculated concentrations and plug into the following ICE table. Let x = [H^{+}] at equilibrium (Do not calculate “x” yet)
Use the K_{a} expression and determine the concentration of H^{+} and the pH of this solution.


Region 3: At the Equivalence Point  
What volume of the titrant has been added to the starting material at the equivalence point for this titration? mL At the equivalence point for a weak acidstrong base titration an equal number of moles of OH^{} and H^{+}have reacted, producing a solution of water and salt. What affects the pH at this point for a strongacid/strongbase titration?
First, fill out the BCA table to determine the moles of material at the equivalence point of the titration. (Fill in actual values, do not use x. Also be sure to use moles)
Use these calculated concentrations and plug into the following ice table. Let x = [OH^{}] at equilibrium (Do not calculate “x” yet)
Use the K_{a} expression and determine the concentration of OH^{} and the pOH of this solution.

Question 1.
Region 1: Initial pH
First, assume the acid:
CH3COOH
to be HA, for simplicity, so it will ionize as follows:
HA <> H+ + A
where, H+ is the proton and A the conjugate base, HA is molecular acid
Ka = [H+][A]/[HA]; by definition
initially
[H+] = 0
[A] = 0
[HA] = 0.187 ;
the change
initially
[H+] = + x
[A] = + x
[HA] =  x
in equilbrirum
[H+] = 0 + x
[A] = 0 + x
[HA] = 0.187  x
substitute in Ka
Ka = [H+][A]/[HA]
Ka = x*x/(Mx)
x^2 + Kax  M*Ka = 0
if M = 0.187 M; then
x^2 + (1.8*10^5)x  0.187 *(1.8*10^5) = 0
solve for x
x =0.001825
substitute
[H+] = 0 + 0.001825= 0.001825M
[A] = 0 + 0.001825= 0.001825M
[HA] = M  x = 0.187 0.001825= 0.185175 M
pH = log(H+) = log(0.001825) = 2.738
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