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):

Answer 1

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   =                        

Answer:

pH = 4.81

pH = 10.40

pH = 12.04

Explanation:

a)

Number of moles of butanoic acid

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $10.00 \ mL * \frac{L}{1000 \ mL }* \frac{0.1000 \ mol }{L}$

= 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] = $\frac{0.001000 \ mol}{ 0.03000 \ L }$

= 0.03333 M

Concentration of [CH₃CH₂COO⁻]  = $\frac{0.001000 \ mol}{ 0.03000 \ L}$

= 0.03333 M

By Henderson- Hasselbalch equation

pH = pKa + log $\frac{conjugate \ base}{acid }$

pH = pKa + log $\frac{CH_3CH_2CH_2COO^-}{CH_3CH_2CH_2COOH}$

PH = 4.81  + log $\frac{0.03333}{0.03333}$

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

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $20.10 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 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 × $\frac{1 \ L}{1000 \ mL }$

= 0.04010 L

Concentration of acid [OH⁻] = $\frac{0.000010 \ mol}{0.04010 \ L }$

= $2.494*10^{-4} M$

Using the ionic  product of water:

$[H_3O^+] = \frac{K \omega }{[OH^-]}$

where

$K \omega = 10^{-14}$

$[H_3O^+] = \frac{1.0*10^{-14}}{2.494*10^{-14}}$

= $4.0*10^{-11}M$

pH = - log $[H_3O^+}]$

pH = - log [$4.0*10^{-11}M$]

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

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $25.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 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 × $\frac{1 \ L}{1000 \ mL }$

= 0.04500 L

Concentration of acid [OH⁻] = $\frac{0.0005000 \ mol}{ 0.04500 \ L }$

= 0.01111 M

Using the ionic product of water $[H_3O^+] = \frac{K \omega }{[OH^+]}$

= $\frac{1.0*10^{-14}}{0.01111}$

= $9.0*10^{-13}$ M

pH = - log $[H_3O^+}]$

pH = - log [$9.0*10^{-13}M$]

pH = 12.04

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