When a solution contains a weak acid and its conjugate base or a weak base and i

Question

When a solution contains a weak acid and its conjugate base or a weak base and its conjugate acid, it will be a buffer solution. Buffers resist change in pH following the addition of acid or base. A buffer solution prepared from a weak acid (HA) and its conjugate base (A) is represented as

HA(aq)H+(aq)+A(aq)

The buffer will follow Le Châtelier's principle. If acid is added, the reaction shifts to consume the added H+, forming more HA. When base is added, the base will react with H+, reducing its concentration. The reaction then shifts to replace H+ through the dissociation of HA into H+ and A. In both instances, [H+] tends to remain constant.

The pH of a buffer is calculated by using the Henderson-Hasselbalch equation:

pH=pKa+log[A][HA]

Part A

What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.608 mol of NaAin 2.00 L of solution? The dissociation constant Ka of HA is 5.66×107.

Part B

What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.

Explanation / Answer

A) pH = pKa+log [A]/ [HA]

    pH = - logKa+log [A]/ [HA]

   pH = - logKa+log [NaA]/ [HA]

Given that volume of solution = 2.00 L

              moles of HA = 0.809 mol

            moles of NaA = 0.608 mol

      Hence, [HA] = moles of HA/ volume of solution = 0.809 mol/ 2.00 L = 0.4045 M

                [NaA] = moles of NaA/ volume of solution = 0.608 mol/ 2.00 L = 0.3040 M

      [HA] = 0.4045 M

     [NaA] = 0.3040 M

Given that Ka of HA = 5.66×107

pH = - logKa+log [NaA]/ [HA]

    = - log (5.66×107) + log (0.3040 M/0.4045 M)

   = 6.123

pH = 6.123

Therefore, pH of the buffer = 6.123

B)

Given that 0.150 mol of HCl is added to the buffer from Part A

NaA            +     HCl              --------------->     HA    + NaCl

0.608 mol          0.150 mol                           0

-------------------------------------------------------------------------------------

0.608-0.150          0                                       0.150 mol

= 0.458 mol

Hence,

moles of HA = initial moles + 0.150 mol = 0.809 mol + 0.150 mol = 0.959 mol

moles of NaA = 0.458 mol

Hence, [HA] = moles of HA/ volume of solution = 0.959 mol/ 2.00 L = 0.4795 M

                [NaA] = moles of NaA/ volume of solution = 0.458 mol/ 2.00 L = 0.2290 M

      [HA] = 0.4795 M

     [NaA] = 0.2290 M

Given that Ka of HA = 5.66×107

pH = - logKa+log [NaA]/ [HA]

    = - log (5.66×107) + log (0.2290 M/0.4795 M)

   = 5.926

pH = 5.926

Therefore, pH of the buffer after addition of 0.150 mol HCl = 5.926

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