Question

Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution:

(Part A) a solution that is 0.170 M in HC2H3O2 and 0.115 M in KC2H3O2. Express your answer using two decimal places.

(Part B) a solution that is 0.195 M in CH3NH2 and 0.135 M in CH3NH3Br. Express your answer using two decimal places.

Answer 1

Clearly in both cases we have to calculate the pH of buffer solutions. In first case, the solution is of a weak acid and its salt, in the second one, the solution is of a weak base and its salt.

(Part A) The pH of a solution of weak acid and its salt is given by Henderson-Hasselbalch equation,

$pH\, =\, pKa\, +\, log\, ([salt]\div [acid])$

where, the [] sign indicates the concentration of the mentioned species.

Now, pKa for the acid given, i.e, acetic acid is 4.74, the concentration of the salt is [salt] = 0.115 M and the concentration of free acid is [acid] = 0.170 M. So according to the equation given,

  $pH\, =\, 4.74\, +\, log\, (0.115\div 0.170)\, =\, 4.57$

(Part B) The pH of a solution of weak base and its salt is given by Henderson-Hasselbalch equation,

$pOH\, =\, pKb\, +\, log\, ([salt]\div[base])$

Now, for CH₃NH₂ the pKa value is 10.63, thus pKb value is (14 - 10.63) = 3.37. The concentration of the salt is [salt] = 0.135 M and the concentration of free base is [base] = 0.195 M. So according to the equation given,

  $pOH\, =\, 3.37\, +\, log\, (0.135\div0.195)\, =\,3.21$

Since, pH= 14 - pOH, thus pH of the solution is (14 - 3.21) = 10.79