Question

A chemistry graduate student is given of a benzoic acid solution. Benzoic acid is a weak acid with . What mass of should the student dissolve in the solution to turn it into a buffer with pH ? You may assume that the volume of the solution doesn't change when the is dissolved in it. Be sure your answer has a unit symbol, and round it to significant digits.

Answer 1

Complete Question

A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with $pKa = 4.2$. What mass of KC6H5CO2 should the student dissolve in the HC6H5CO2 solution to turn it into a buffer with pH =4.63? You may assume that the volume of the solution doesn't change when the KC6H5CO2 is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.

Answer:

The mass would be  $mass \ of \ KC_6H_5CO_2 = 54 g$

Explanation:

   From the question we are told that

           concentration of benzoic acid is $M_B = 1M$

            The volume of benzoic acid is $V_B = 125mL = 0.125L$

            The value of the experimental parameter is $pKa = 4.2$

          The value of PH is $PH =4.63$

Generally the number of moles is mathematically represented as

             No of moles = concentration * volume  

                            $N = M_B * V_B$

Substituting values

                          $N = 1 * 0.125 = 0.125\ moles$

 Generally PH is mathematically represented as

               $PH = pKa + log\frac{[KC_6H_5CO_2]}{[HC_6H_5CO_2]}$

Where $[KC_6H_5CO_2]$ is the No of moles of $KC_6H_5CO_2$ and $[HC_6H_5CO_2]$ is the number of moles of benzoic acid

              $4.3 = 4.2 + log\frac{[KC_6H_5CO_2]}{0.125}$

             $4.63 - 4.2 = log \frac{[KC_6H_5CO_2]}{0.125}$

                           $0.43 =log \frac{[KC_6H_5CO_2]}{0.125}$

                         $10^{0.43} =\frac{[KC_6H_5CO_2]}{0.125}$

                         $2.6915 *0.125 = [KC_6H_5CO_2]$

                          $[KC_6H_5CO_2]= 0.3364$

Generally the number of moles is mathematically represented as

              $No \ of \ moles \ = \frac{Mass}{Molar\ Mass}$

For $[KC_6H_5CO_2]$

             $0.3364 = \frac{mass \ of \ KC_6H_5CO_2}{Molar \ Mass \ of \ KC_6H_5CO_2 }$

         Molar mass of $KC_6H_5CO_2$ = 160g/mole

         $0.3364 * 160 =mass \ of \ KC_6H_5CO_2$

         $mass \ of \ KC_6H_5CO_2 = 54 g$