Posted by Cassandra on Monday, April 2, 2012 at 11:39pm.
You want 25 mL of 0.1M; that is mols = g/molar mass = ?
You want to convert half of that to the acid; therefore, 1/2 x mols tris = ?
Now, M HCl = mols HCl/L HCl.
You kow M and mols, solve for L HCl and convert to mL.
A chemist needs to make 50 mL buffer solution with a pH value of 7.25 using the acid/conjugate base pair of KH2PO4 and Na2HPO4 (which has an effective pKa = 6.86). The buffer solution needs to be made so that the total concentration of buffer is 0.10 M (that is, [acid] + [base] = 0.10 M). Calculate the mass of KH PO and Na HPO needed using the following formula:
10^(7.25−𝑝𝐾𝑎) = (0.10 M×0.050 L−mol HA)/mol HA
Solve for mol acid and then use the molecular weight of the acid to calculate the mass (grams) of acid. Then solve for mol base using mol A- + mol HA = 0.10 M × 0.05 L and calculate the mass (grams) of base.
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.