Describe how you would prepare the following aqueous solutions?

You have to prepare a pH 5.06 buffer, and you have the following 0.10M solutions available...?

• Answer:

  The pH of the buffer is given by the Henderson - Hasselbalch equation: pH = pKa + log ([salt]/[acid]) pKa for CH3COOH = 4.74 Substitute into the H-H equation 5.06 = 4.74+ 0.32 Therefore log ([salt]/[acid]) = 0.32 [salt]/[acid] = 10^0.32 [salt]/[acid] = 2.089 = 2.1 The molar ratio salt : acid must be 2.1 CH3COONa : 1.0CH3COOH Because the molar concentrations of the two components are equal , we can use volumes: You want a final volume 1000mL Volume of CH3COONa = 2.1/3.1*1000 = 677.4 mL Volume of CH3COOH = 1.0/3.1*1000 = 322.6mL Always a good idea to check : Mol CH3COONa in 677.4mL of 0.1 M solution = 677.4/1000*0.1 = 0.06774 Mol CH3COOH in 322.6 mL of 0.1M solution = 0.03226 pH = pKa + log ([salt]/[acid]) pH = 4.74 + log (0.06674/0.03226) pH = 4.74 + log 2.0688 pH = 4.74 + 0.32 pH = 5.06 which is what you want.