Question

A 25.00 mL sample of a 0.100 M trimethylamine (Kb = 6.46 x 105) is titrated with 0.125 M solution of HCl. What is the pH of the solution after 10.0 and 20.0 mL of acid have been added?

Answer 1

Answer :

(1) The pH of the solution after addition of 10.0 mL acid is, 4.19

(2) The pH of the solution after addition of 20.0 mL acid is, 10.77

Explanation :

(1) To calculate the pH of the solution after addition of 10.0 mL of acid.

First we have to calculate the moles of $(CH_3)_3N$ and $HCl$.

$\text{Moles of }(CH_3)_3N=\text{Concentration of }(CH_3)_3N\times \text{Volume of solution}=0.100M\times 0.025L=0.0025mole$

$\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}=0.125M\times 0.010L=0.00125mole$

In this, HCl is limiting reactant because it is present in less amount.

The balanced chemical reaction is,

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

From the reaction we conclude that the mole ratio of $(CH_3)_3N,HCl\text{ and }(CH_3)_3N^+Cl^-$ are 1 : 1 : 1.

Moles of $(CH_3)_3N$ left = Initial moles of $(CH_3)_3N$ - Moles of $HCl$ added

Moles of $(CH_3)_3N$ left = 0.0025 - 0.00125 = 0.00125 mole

Moles of $(CH_3)_3N^+Cl^-$ = Moles of HCl = 0.00125 mole

Now we have to calculate the $pK_b$

$pK_b=-\log (6.46\times 10^{5})$

$pK_b=4.19$

Now we have to calculate the pOH by using Henderson-Hasselbalch equation.

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

Total volume of solution = 25 + 10 = 35 mL = 0.035 L

Concentration of $(CH_3)_3N$ = $\frac{Moles}{Volume}=\frac{0.00125mole}{0.035L}=0.0357M$

Concentration of $(CH_3)_3N^+Cl^-$ = $\frac{Moles}{Volume}=\frac{0.00125mole}{0.035L}=0.0357M$

Now put all the given values in this expression, we get:

$pOH=4.19+\log (\frac{0.0357}{0.0357})$

$pOH=4.19$

Now we have to calculate the pH of the solution.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.19=9.81$

The pH of the solution after addition of 10.0 mL acid is, 9.81

(2) To calculate the pH of the solution after addition of 20.0 mL of acid.

First we have to calculate the moles of $(CH_3)_3N$ and $HCl$.

$\text{Moles of }(CH_3)_3N=\text{Concentration of }(CH_3)_3N\times \text{Volume of solution}=0.100M\times 0.025L=0.0025mole$

$\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}=0.125M\times 0.020L=0.00025mole$

In this, HCl is limiting reactant because it is present in less amount.

The balanced chemical reaction is,

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

From the reaction we conclude that the mole ratio of $(CH_3)_3N,HCl\text{ and }(CH_3)_3N^+Cl^-$ are 1 : 1 : 1.

Moles of $(CH_3)_3N$ left = Initial moles of $(CH_3)_3N$ - Moles of $HCl$ added

Moles of $(CH_3)_3N$ left = 0.0025 - 0.00025 = 0.00225 mole

Moles of $(CH_3)_3N^+Cl^-$ = MOles of HCl = 0.00025 mole

Now we have to calculate the pOH by using Henderson-Hasselbalch equation.

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

Total volume of solution = 25 + 20 = 45 mL = 0.045 L

Concentration of $(CH_3)_3N$ = $\frac{Moles}{Volume}=\frac{0.00225mole}{0.045L}=0.05M$

Concentration of $(CH_3)_3N^+Cl^-$ = $\frac{Moles}{Volume}=\frac{0.00025mole}{0.045L}=0.0055M$

Now put all the given values in this expression, we get:

$pOH=4.19+\log (\frac{0.0055}{0.05})$

$pOH=3.23$

Now we have to calculate the pH of the solution.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-3.23=10.77$

The pH of the solution after addition of 20.0 mL acid is, 10.77