posted by Lucy
Starting out with 50 mL of 0.20 M NaHCO3, calculate how many mL of 0.50 M NaOH solution to add to make 100 mL of approximately 0.10 M (total) buffer solution with a pH of 10.35. By adding NaOH, some of the NaHCO3 gets converted to the conjugate Na2CO3. This conjugate acid/base pair has an effective pKa of 10.00. Calculate mL of 0.50 M NaOH needed from the following calculations:
10.35-pK =log[(mol CO3^2-)/(mol HCO3^-)]
There is initially Na2CO3 present in the 50 mL of 0.20 M NaHCO3 solution. By adding some NaOH, you will create some Na2CO2 in the solution and also decrease the mol of NaHCO2. In fact, mol NaOH added = mol Na CO in buffer solution. Therefore,
10^(10.35−𝑝𝐾𝑎)= [(mol NaOH added)/(0.20 M×0.050 L−mol NaOH added)]
Solve for mol NaOH added in the above equation. Then, use the concentration of the NaOH (0.50 mol/L) to calculate the calculate the mL of NaOH necessary to prepare the buffersolution.