Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M trimethylamine, (CH3)3N(aq), with 0.2000 M HCl(aq) after 7.5 mL of the acid have been added. Kb of trimethylamine = 6.5 x 10-5.

Answer 1

Answer: The pH of the solution is 9.33

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$    .....(1)

• For HCl:

Molarity of HCl solution = 0.200 M

Volume of solution = 7.5 mL

Putting values in equation 1, we get:

$0.200M=\frac{\text{Moles of HCl}\times 1000}{7.5}\\\\\text{Moles of HCl}=\frac{0.200\times 7.5}{1000}=0.0015mol$

• For trimethylamine:

Molarity of trimethylamine solution = 0.100 M

Volume of solution = 20 mL

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of trimethylamine}\times 1000}{20}\\\\\text{Moles of trimethylamine}=\frac{0.100\times 20}{1000}=0.002mol$

The chemical reaction for trimethylamine and HCl follows the equation:

                    $(CH_3)_3N+HCl\rightarrow (CH_3)_3NH^++Cl^-$

Initial:             0.002      0.0015

Final:               0.0005          -         0.0015

Total volume of the solution = [20 + 7.5] = 27.5 mL = 0.0275 L   (Conversion factor:  1 L = 1000 mL)

To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})$

$pOH=pK_b+\log(\frac{[(CH_3)_3NH^+]}{[(CH_3)_3N]})$

where,

$pK_b$ = negative logarithm of base dissociation constant of trimethylamine = 4.19

$[(CH_3)_3NH^+]=\frac{0.0015}{0.0275}$

$[(CH_3)_3NH]=\frac{0.0005}{0.0275}$  

pOH = ?

Putting values in above equation, we get:

$pOH=4.19+\log(\frac{(0.0015/0.0275)}{(0.0005/0.0275)})\\\\pOH=4.67$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-4.67=9.33$

Hence, the pH of the solution is 9.33