You prepare a buffer solution by dissolving 2.00 g each of benzoic acid, C6H5COOH, and

Question

sodium benzoate, NaC6H5COO, in 750.0 mL water. (2 pts each)
a) What is the pH of this buffer? Assume that the solution’s volume is 750.0 mL.
b) If 0.55 mL of 12 M HCl is added to 0.750 L of this buffer solution, what will be the pH
of the resulting solution?
c) If 0.35 mL of 15 M NH3 is added to 0.750 L of this buffer solution, what will be the pH
of the resulting solution?

Answer 1

My calculations are estimates. You should recalculate everything that follows.
a. Let's call benzoic acid, C6H5COOH, HBz and sodium benzoate, NaBz.
HBz molar mass = 122, NaBz molar mass = 144
mols HBz = 2/122 = about 0.016
(HBz) = 0.016/0.750 = ? M
mols NaBz = 2/144 = about 0.014
(NaBz) = 0.014/0.750 = ? M
pH = pKa + log (NaBz)/(HBz) = ?
b. I prefer to wor in millimoles for this. You can convert to M if you wish.
millimoles HBz initially = about 16. Remember that's just an estimate.
millimoles NaBz = 14
HCl added = 0.55 mL x 12 M HCl = 6.6 millimoles
.....................Bz^- + H^+ ==> HBz
Initial............14........0..............16
add.........................6.6.................
change........-6.6....-6.6.............+6.6
equilibrium..7.4.........0...............6.6

The equilibrium line is millimoles you have in the final solution. (Bz^-) = millimoles/mL = 7.4/750 =? M
M HBz = 6.6/750 = ? M
Plug these into the Henderson-Hasselbalch equation and solve for pH.

c. Same thing as b except it's base being added and not acid.
You want to start with HBz + NH3 ==> NH4Bz
Post your work if you get stuck.