Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 × 10 − 5), with 0.1000 M NaOH solution after the following additions of titrant (total volume of added base given):

Here is the full question

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 × 10 − 5), with 0.1000 M NaOH solution after the following additions of titrant (total volume of added base given):

a) 10.00 mL  

pH   =                        

b) 20.10 mL

pH   =                        

c) 25.00 mL

pH   =                        

pH = 4.81

pH = 10.40

pH = 12.04

Explanation:

a)

Number of moles of butanoic acid

=

= 0.002000 mol

Number of moles of NaOH added

=

= 0.001000 mol

pKa of butanoic acid = - log Ka

= - log ( 1.54 × 10⁻⁵)

= 4.81

Equation for the reaction is expressed as follows:

CH₃CH₂CH₂COOH    +  OH⁻   ----->   CH₃CH₂COO⁻   +   H₂O

The ICE Table is expressed as follows:

                    CH₃CH₂CH₂COOH    +  OH⁻   ----->   CH₃CH₂COO⁻   +   H₂O

Initial                 0.002000                  0.001000               0

Change            - 0.001000                - 0.001000         + 0.001000  

Equilibrium         0.001000                         0                   0.001000

Total Volume = (20.00 + 10.00 ) mL

=  30.00 mL = 0.03000 L

Concentration of  [CH₃CH₂CH₂COOH] =

= 0.03333 M

Concentration of [CH₃CH₂COO⁻]  =

= 0.03333 M

By Henderson- Hasselbalch equation

pH = pKa + log

pH = pKa + log

PH = 4.81  + log

pH = 4.81

Thus; the pH of the resulted buffer solution after 10.00 mL of NaOH was added = 4.81

b )

After the equivalence point, we all know that the pH of the solution will now definitely be determined by the excess H⁺

Number of moles of butanoic acid

=

= 0.002000 mol

Number of moles of NaOH added

=

= 0.002010 mol

Following the previous equation of reaction , The ICE Table for this process is as follows:

                    CH₃CH₂CH₂COOH    +  OH⁻   ----->   CH₃CH₂COO⁻   +   H₂O

Initial                 0.002000                  0.002010               0

Change           - 0.002000                -0.002000         + 0.002000  

Equilibrium         0                                0.000010            0.002000

We can see here that the base is present in excess;

NOW, number of moles of base present in excess

= ( 0.002010 - 0.002000) mol

= 0.000010 mol

Total Volume = (20.00 + 20.10 ) mL

= 40.10 mL ×

= 0.04010 L

Concentration of acid [OH⁻] =

=

Using the ionic  product of water:

where

=

pH = - log

pH = - log []

pH = 10.40

Thus, the pH of the solution after the equivalence point = 10.40

c)

After the equivalence point, pH of the solution is determined by the excess H⁺.

Number of moles of butanoic acid

=

= 0.002000 mol

Number of moles of NaOH added

=

= 0.002500 mol

From our chemical equation; The ICE Table can be illustrated as follows:

                    CH₃CH₂CH₂COOH    +  OH⁻   ----->   CH₃CH₂COO⁻   +   H₂O

Initial                 0.002000                 0.002500               0

Change           - 0.002000                -0.002000           +0.002000  

Equilibrium         0                               0.000500            0.002000

Base is present in excess

Number of moles of base present in excess = [ 0.002500 - 0.002000] mol

= 0.000500 mol

Total Volume = ( 20.00 + 25.00 ) mL

= 45.00 mL

= 45.00 ×

= 0.04500 L

Concentration of acid [OH⁻] =

= 0.01111 M

Using the ionic product of water

=

= M

pH = - log

pH = - log []

pH = 12.04

Thus, the pH of the solution after the equivalence point = 12.04

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explanation:

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