Question

(1) Write the net ionic equation for the reaction that occurs when equal volumes of 0.233 M aqueous acetylsalicylic acid (aspirin) and aniline are mixed. It is not necessary to include states such as (aq) or (s).

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

 1)  The ionic equation is  

$HC_9H_7O_4 + (C_2 H_5)_3 N -----> C_9H_7O_4^- + (C_2 H_5)_3 NH^-$

2

  At equilibrium The  aniline will be  favored

3

The pH  of the solution is   $pH = 7.12$

Explanation:

  From the question we are told that

      The concentration of aspirin is  $C_A = 0.233M$

      The acid dissociation  constant for aspirin is  $K_a = 3.0*10^{-4}$

       The base  dissociation constant for aniline is  $K_b = 7.4 *10^{-10}$

The molecular formula for aspirin  is  $HC_9H_7O_4$

The molecular formula for aniline  is   $(C_2 H_5)_3 N$

So the net ionic reaction is  

     $HC_9H_7O_4 + (C_2 H_5)_3 N -----> C_9H_7O_4^- + (C_2 H_5)_3 NH^-$

Generally pKa is mathematically evaluated as

                  $pKa = -log(3.0*10^{-4})$

                           $= 3.52$

Generally pKb is mathematically represented as

                  $pKa = -log(7.4*10^{-10 })$

                           $= 9.13$

Generally

                    pKa + pKb = 14

So for the triethylammonium salt  produced the pKa is

                  $pK_a__{s}} = 14 - 3.28$

                            $= 10.72$          

 As a result of the   pKa  of  aspirin being lower that that of  triethylammonium salt at equilibrium it implies that  aniline would be favored

The pH of the solution is mathematically represented as

           Since they are of equal mole

     $pH = \frac{1}{2} (pKa + pKb + pKw)$

Where pKw is the pKw of water which has a value of 14

$pH = 0.5(14 + 3.52 + 3.28)$

    $pH = 7.12$

Answer 2

Answer:

Check the explanation

Explanation:

Ka for aspirin = $3.0 X 10^{-4}$

Therefore,

pKa = - log($3.0 X 10^{-4}$) = 3.52

Kb for Triethylamine = 5.2 x 10-4

Therefore,

pKb = -log($5.2 X 10^{-4}$) = 3.28

We have,

pKa + pKb = 14

pKa = 14 – pKb = 10.71

pKa for the triethylammonium salt = 10.71

The net ionic reaction can be written as,

$HC_9H_7O_4 + (C_2H_5)3_N$      ⇒      $C_9H_7O_4- + (C_2H_5)_3NH+$

Since pKa of Aspirin is lower (3.52) than that of triethylammonium salt (10.71), at equilibrium triethyl amine will be favored.

pH of equimolar solution can be calculated as,

pH = ½ (pKw + pKa –pKb) = ½ (14 + 3.52 – 3.28) = 7.12

pH of resulting solution = 7.12