If you add 2.50 mL of .150 M HCl to 100 mL of a buffer consisting of .100 M CH_3COOH and .200 M CH_3COONa, what will be the change in pH? (Ka of CH_3COOH is 1.7 x 10^-5).

Calculate pH from the Henderson-Hasselbalch equation to give you the initial pH.

Then to that buffer you add 2.50 mL x 0.150 M HCl to 100 mL. That will add 0.000375 moles HCl. That will increase the mole CH3COOH by that amount and decrease moles CH3COONa by that amount. Calculate new values for concn CH3COOH and CH3COONa, plug into the HH equation and calculate the new pH.

To find the change in pH when adding an acid to a buffer solution, we can use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentrations of the acid and its conjugate base:

pH = pKa + log([conjugate base]/[acid])

First, let's determine the initial concentration of the acid (CH3COOH) and its conjugate base (CH3COONa) in the buffer solution before any reaction:

Initial [CH3COOH] = 0.100 M
Initial [CH3COONa] = 0.200 M

We also need to determine the volume of the buffer solution:

Volume of buffer solution = 100 mL

Now, let's calculate the moles of HCl added:

Moles of HCl = (concentration of HCl) x (volume of HCl added)
= 0.150 M x 0.00250 L
= 0.000375 moles

Since the HCl reacts with the CH3COONa in a 1:1 ratio, the same amount of HCl will react with the CH3COO- ions, resulting in a decrease in the concentration of the conjugate base.

Concentration of [CH3COO-] after reaction = Initial [CH3COONa] - moles of HCl reacted
= 0.200 M - 0.000375 moles / (0.100 L + 0.00250 L)
≈ 0.198 M

Now we can use the Henderson-Hasselbalch equation to calculate the change in pH.

ΔpH = pHfinal - pHinitial
= (pKa + log([conjugate base]/[acid]))final - (pKa + log([conjugate base]/[acid]))initial
= (pKa + log(0.198/0.100)) - (pKa + log(0.200/0.100))
= log(0.198/0.100) - log(0.200/0.100)
= log(0.198/0.200)
≈ log(0.99)
≈ -0.004

Therefore, the change in pH when adding 2.50 mL of 0.150 M HCl to 100 mL of the buffer solution will be approximately -0.004.