Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added. What is the pKa of your unknown acid?

Answer 1

Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is $pK_a =7.82$

Explanation:

From the question we are told

    The volume of base is  $V_B = 26.mL = 0.0260L$

     The pH of solution is  $pH = 7.82$

      The concentration of the acid is $C_A = 0.1M$

From the pH we can see that the titration is between a strong base and a weak acid

 Let assume that the the volume of acid is  $V_A = 18mL= 0.018L$

Generally the concentration of base

                    $C_B = \frac{C_AV_A}{C_B}$

Substituting value  

                     $C_B = \frac{0.1 * 0.01800}{0.0260}$

                    $C_B= 0.0692M$

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

           $HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}$

Now before the reaction the number of mole of base is  

            $No \ of \ moles[N_B] = C_B * V_B$

Substituting value  

                    $N_B = 0.01300 * 0.0692$

                         $= 0.0009 \ moles$    

                                 

Now before the reaction the number of mole of acid is  

            $No \ of \ moles = C_B * V_B$

Substituting value  

                    $N_A = 0.01800 *0.1$

                         $= 0.001800 \ moles$

Now after the reaction the number of moles of  base is  zero  i.e has been used up

    this mathematically represented as

                         $N_B ' = N_B - N_B = 0$

    The  number of moles of acid is  

             $N_A ' = N_A - N_B$

                   $= 0.0009\ moles$

The pH of this reaction can be mathematically represented as

                 $pH = pK_a + log \frac{[base]}{[acid]}$

Substituting values

                  $7.82 = pK_a +log \frac{0.0009}{0.0009}$

                  $pK_a =7.82$