Calculating the change in pH.....HHHEEELLLPPP?

A beaker with 130 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 6.60 mL of a 0.330 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.Calculating the change in pH.....HHHEEELLLPPP?
Use the Henderson-Hasselbalch equation to find the ratio of acid and conjugate base the current solution has:



pH = pKa + log[A-/HA]



[HA] = CH3COOH

[A-] = CH3COONa



pH = 5.00, pKa = 4.760



5.00 = 4.760 = log[A-/HA]



1.74 = [A-]/[HA]



1.74[HA] = [A-]; this is the ratio between acid and conjugate base.



The total molarity of the solution is 0.100, so the total molarity of [HA] and [A-] has to equal to 0.100.

[HA] + [A-] = 0.100



Plug in the equation from the top:

2.74[HA] = 0.100



[HA] = 0.0365 M

[A-] = 0.0635 M



Multiply by the volume to find the mmoles of each you have.



[HA] = 0.0365 * 130 mL = 4.475 mmoles HA

[A-] = 0.0635 * 130 mL = 8.255 mmoles A-



6.60 mL * 0.330 M = 2.178 mmoles HCl = 2.178 mmoles H+ ions



HCl reacts completely with A- to produce HA, so you will lose 2.178 mmoles of A- and produce 2.178 mmoles HA.



4.475 + 2.178 = 6.653 mmoles HA

8.255 - 2.178 = 6.077 mmoles A-



Stick it back into the Henderson-Hasselbalch equation. You could divide by the total volume or leave it in mmoles, but it doesn't matter since it's a ratio of the two:



pH = pKa + log[A-/HA]

pH = 4.760 + log[6.077/6.653] = 4.76 - 0.04 = 4.72

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