Question

POST-LABORATORY QUESTIONS 1. Explain how it is possible for the concentration of acetic acid to be higher than that of hydrochloric acid in solutions that have the same H3O ion concentration (same pH)? Remember, both acids ionize to form only one mole of hydronium ions per mole of acid that ionizes. 2. Do you think it is accurate to consider the titration end point as the same as the "equivalence point"? Why? How do you know that an indicator's color change depends upon the solution's pH and not on the kind of acid and base being titrated or the volume of titrant added? 3. During the titration of acid and base, does the "equivalence point" always occur at pH-7.0? Explain when the equivalence point occurs at pH=7 and when it does not. 4.

Answer 1

Following is the - complete Answer -&- Explanation: for the first question( i.e. Question - 1 ), of the given: Question Set.....in...typed format...

$\Rightarrow$Answer:

The concentration of acetic acid ( CH₃COOH ): is higher than: that of hydrochloric acid ( HCl ) , when solutions of both the acids, are having the same value of pH : is because of the fact, that: the value of the acid dissociation constant ( K_a ) ; of  HCl solution, is much much higher than the value of K_a of CH₃COOH solution...[ i.e. as it is vividly discussed below: under: Step - 6 ]

$\Rightarrow$Explanation:

Following is the complete Explanation: for the above: Answer...

• Step - 1:

We know the following values of acid dissociation constant: of the given: acids, as the following:

1. For   HCl solution: the value of K_a = 1.3 x 10⁶  
2. For CH₃COOH : the value of K_a = 1.8 x 10^-5

• ​​​​​​​Step - 2:

​​​​​​​Let's consider the balanced chemical equations: of the dissociation of the above aqueous solutions, of the given: acids: into ions:

1. HCl (aq) + H₂O (l)   $\rightleftharpoons$ H₃O⁺ (aq) + Cl ^- (aq)   ------------------------------Equation - 1
2. CH₃COOH (aq) + H₂O (l)   $\rightleftharpoons$   H₃O⁺ (aq) +   CH₃COO ^- (aq) ------------------Equation - 2

• ​​​​​​​Step - 3:

​​​​​​​Let's consider the concentration of H₃O⁺ ions: of both the aqueous solutions; of the given acids, to be = ' x ' M ( mol/L ) , and let's form the ICE Tables for both the acids, based on Equation - 1 and Equation - 2.

• Step - 4:

​​​​​​​Following is the ICE Table formed on the basis of Equation - 1:

	[HCl]	[H3O+]	[Cl - ]
Initial( concentration)	Y	0.0	0.0
Change (concentration)	- X	+X	+X
Equilibrium (concentration)	Y - X	+X	+X

$\Rightarrow$Therefore: for  HCl

  K_a = 1.3 x 10⁶  = X² / ( Y - X ) -----------------------Equation - 3

• Step - 5:

​​​​​​​Following is the ICE Table formed on the basis of Equation - 2:

	[CH3COOH]	[H3O+]	[CH3COO - ]
Initial( concentration)	Z	0.0	0.0
Change (concentration)	- X	+X	+X
Equilibrium (concentration)	Z - X	+X	+X

$\Rightarrow$Therefore: for  CH₃COOH

  K_a = 1.8 x 10^-5= X² / ( Z - X ) ---------------------------Equation - 4

[ Note:  The concentration of H₃O⁺  = X (mol/L ); for both the acids ​​​​​​​, at equilibrium...]

• Step - 6:

​​​​​​​If we divide: Equation - 3 : by  Equation - 4 : We will get the following:

$\Rightarrow$ ( Z - X ) / ( Y - X ) = 1.3 x 10⁶/ 1.8 x 10^-5 = 7.22 x 10¹⁰   >>  1

$\Rightarrow$ ( Z - X ) >> ( Y - X )

$\Rightarrow$   i.e. Z >> Y ; i.e.  [CH₃COOH] >> [HCl ] ; i.e. concentration of CH₃COOH >> concentration of HCl