Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M trimethylamine, (CH3)3N(aq), with 0.2000 M HClO4(aq) after 9.48 mL of the acid have been added.

Answer 1

Complete Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M trimethylamine, (CH3)3N(aq), with 0.2000 M HClO4(aq) after 9.48 mL of the acid have been added.Kb of trimethylamine = 6.5 x 10-5.

Answer:

The pH is  $pH = 9.84$  

Explanation:

From  the question we are told that

   The  volume of trimethylamine, (CH3)3N(aq) is $V_{t} = 20.00mL$

    The concentration of trimethylamine is $C_t = 0.1000 \ M$  

    The  volume of HClO4(aq) is $V_{h} = 9.48 mL$

    The  concentration of  HClO4(aq) is  $C_h = 0.200 M$

      The Kb  value is  $K_b = 6.5 * 10^{-5}$

Generally the the pOH of this reaction is mathematically represented as

       $pOH = pK_b + log [\frac{N_h}{N_b} ]$

Here $N_h$ is the number of moles of acid which is evaluated as

      $N_h = C_h * V_h$

=>    $N_h = 0.200 * 9.48$

=>    $N_h = 1.896$

Here $N_t$ is the number of moles of acid which is evaluated as

      $N_t = C_t * V_t$

=>    $N_t = 0.100 * 20$

=>    $N_t = 2$

So

     $pOH = -log(K_b) + log [\frac{N_h}{N_b} ]$

     $pOH = -log(6.5*10^{-5}) + log [\frac{1.896}{2} ]$

=>   $pOH = 4.1639$

Generally the pH is mathematically represented as

       $pH = 14 - pOH$

=>    $pH = 14 - 4.1639$  

=>    $pH = 9.84$