Question

Assume that the reaction used to prepare the buffer goes to completion (i.e., that all the strong base reacts) and use the data in Table A to calculate the initial concentration of the weak acid and its conjugate base in each of the three buffer solutions.

TABLE A

Buffer	mL of 0.50 M weak acid	mL of 0.25 M NaOH	mL of H2O
1	50.00	50.00	0.00
2	50.00	35.00	15.00
3	50.00	25.00	25.00

Answer 1

Ans. Balanced reaction:        AH + NaOH ------------> Na⁺ + A^- + H₂O

Assuming the weak acid to be monoprotic, 1 mol weak acid is neutralized by 1 mol NaOH.

So, the number of moles of NaOH consumed in neutralization reaction is equal to the number of moles of AH neutralized and moles of A^- formed.         

# Buffer 1: Moles of AH = Molarity x Volume of solution in liters

                                                = 0.50 M x 0.050 L

                                                = 0.025 mol

Moles of NaOH = 0.25 M x 0.050 L = 0.0125 mol

Moles of AH neutralized = Moles of A^- formed = moles of NaOH = 0.0125 mol

Remaining moles of AH = Initial moles – Moles neutralized

= 0.025 mol – 0.0125 mol = 0.0125 mol

# Final volume of buffer = 50.0 mL (buffer) + 50.0 mL (NaOH) = 100.0 mL = 0.100 L

Now, concentrations after reaction is complete-

            [AH] = 0.0125 mol / 0.100 L = 0.125 M

            [A^-] = 0.0125 mol / 0.100 L = 0.125 M

# Buffer 2: Moles of AH = 0.50 M x 0.050 L = 0.025 mol

Moles of NaOH = 0.25 M x 0.035 L = 0.00875 mol

Moles of AH neutralized = Moles of A^- formed = moles of NaOH = 0.00875 mol

Remaining moles of AH = Initial moles – Moles neutralized

= 0.025 mol – 0.00875 mol = 0.01625 mol

# Final volume of buffer = 50.0 mL + 35.0 mL + 15.0 mL = 100.0 mL = 0.100 L

Now, concentrations after reaction is complete-

            [AH] = 0.01625 mol / 0.100 L = 0.1625 M

            [A^-] = 0.00875 mol / 0.100 L = 0.0875 M

# Buffer 3: Moles of AH = 0.50 M x 0.050 L = 0.025 mol

Moles of NaOH = 0.25 M x 0.025 L = 0.00625 mol

Moles of AH neutralized = Moles of A^- formed = moles of NaOH = 0.00625 mol

Remaining moles of AH = Initial moles – Moles neutralized

= 0.025 mol – 0.00625 mol = 0.01875 mol

# Final volume of buffer = 50.0 mL + 25.0 mL + 25.0 mL = 100.0 mL = 0.100 L

Now, concentrations after reaction is complete-

            [AH] = 0.01875 mol / 0.100 L = 0.1875 M

            [A^-] = 0.00625 mol / 0.100 L = 0.0625M