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