Posted by **Pearl** on Tuesday, August 2, 2011 at 4:25am.

calculate Ph occuring when 100ml of 0.10m acid is added to 500ml of 0.08m acetate buffer,pka 4.75,ph 4.5

- Biochemistry -
**DrBob222**, Tuesday, August 2, 2011 at 3:01pm
I caution you that m means molality. M means molarity. I assume you meant M.

Use the Henderson-Hasselbalch equation.

a = acid; b = base.

pH = pKa + log(b/a)

4.5 = 4.75 + log b/a

solve for b/a and I get approximately 0.6 (you should do it more accurately).

You have two unknowns and you need a second equation. That one is

a + b = 0.08

Solve the two equations simultaneously. I get a = about 0.05M and b = about 0.03M.

Now you have 500 mL of the buffer so you have how many millimoles of a and b?

500 x 0.05 = about 25 mmoles acid.

500 x 0.03 = about 15 mmoles base.

...........Ac^- + H^+ ==> HAc

initial....15.....0.......25

add............10(100 x 0.1 = 10 mmol)

change....-10....-10......+10

equil......5......0........35

Again, these are approximate; you need to go through the calculations more accurately.

Then plug these equilibrium concns into the HH equation and solve for pH.

I arrived at 3.84.

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