Using a 0.25 M phosphate buffer with a pH of 6.6, you add 0.71 mL of 0.51 M NaOH to 49 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

I just don't understand how to do this and what the answers are.

Question

Using a 0.25 M phosphate buffer with a pH of 6.6, you add 0.71 mL of 0.51 M HCl to 49 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

Using a 0.25 M phosphate buffer with a pH of 6.6, you add 0.71 mL of 0.51 M NaOH to 49 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

I just don't understand how to do this and what the answers are.

DrBob222 · Accepted Answer

You use the Henderson-Hasselbalch equation. Each of these problems has two parts to it. First, you must use the HH equation to determine the (acid) and (base). Second, you plug in the added (H^+) from the HCl and calculate the new pH from the HH equation for problem 1 OR plug in the added (OH^-) from the NaOH and calculate the new pH from the HH equation for problem 2. Here is the outline for problem 1.
pH = pK2 + (B)/(A) where (B) = (base) and (A)-= (acid)
6.6 = 7.20 + B/A. You should look up pK2 for H3PO4 in your text and use that. The 7.20 is from my references.
Solve for B/A. That's two unknowns so you need another equation. You get that from the problem. It says that (A) + (B) = 0.25 M. Solve these two equations simultaneously for (A) and (B). That part isn't chemistry, just math. Post your work if you get stuck. For your reference, I obtained about 0.2M for acid and 0.05M for base. Then you go to part two for the first problem.

I prefer to work in millimoles from here. You have
millimoles buffer = M x mL = 0.25M x 49 mL = 12.25
millimoles base = M x mL = 0.05 x 49 mL = 2.45 = (HPO4^-2)
millimoles acid = M x mL -= 0.20 x 49 mL = 9.8 = (H2PO4^-)
millimoles HCl added = M x mL = 0.51M x 0.71 mL = 0.36
Set up an ICE chart as follows:
............HPO4^2- + H^+ = H2PO4^-
I...........2.45...........0.36........9.80
add .......................0.36..................
C........-0.36...........-0.36........+0.36
E.........2.09................0.........1.02
Take the E values in millimoles, convert to concentrations (M = millimoles/mL) and plug into the HH equation to calculate the new pH of the solution.

Follow the same steps for the added NaOH.

Post all of your work if you get stuck.

Julie · Answer

Where are you getting the 1.02 from for the equilibrium