What is the pH of final solution?

A beaker with 185 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M . A student adds 7.20 mL of a 0.300 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.



Express your answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased.What is the pH of final solution?
First use the Henderson-Hasselbalch equation to calculate the concentrations of acetic acid and acetate in the initial buffer:



pH = pKa + log [acetate]/[acetic acid]

5.00 = 4.76 + log [acetate]/[acetic acid]

log [acetate]/[acetic acid] = 0.24

[acetate]/[acetic acid] = 1.74

[acetate] = 1.74 [acetic acid] **



Now, you know that [acetic acid] + [acetate] = 0.100 M. It's often easier to work in just moles rather than in actual molar concentrations, so multiplying the total concentration by the volume of the buffer solution, you have:

acetic acid + acetate = 0.0185 moles



Substituting the starred equation into the second one gives:

acetic acid + 1.74 acetic acid = 0.0185

2.74 acetic acid = 0.0185

moles acetic acid = 6.75 X 10^-3 moles and then moles acetate = 0.0185 - 6.75 X 10^-3 = 0.01175 moles.



Now, if you add 7.20 mL of 0.300 M HCl, you have added 2.16 X 10^-3 moles of HCl.



When HCl is added to a buffer, the moles of acetate will decrease by exactly the amount of HCl added, and the moles of acetic acid will increase by that same amount. So, after the HCl addition,



moles acetic acid = 6.75 X 10^-3 + 2.16 X 10^-3 = 8.91 X 10^-3 moles

moles of acetate = 0.01175 - 2.16 X 10^-3 = 9.59 X 10^-3 moles.



Now, use the H-H equation to calculate the final pH:



pH = 4.760 + log 9.59X 10^-3 / 8.91 X 10^-3 = 4.79.



So, the pH dropped by 0.21 pH units.