I hope this makes a little sense? Review your acid and base chapters. Ive attached the autonionization of water and my work.
The solution to the problem is as follows:
This is mostly stoichiometry with a little bit of substitution into the acid dissociation equation.
You need intial moles of acid and salt = 0.025 x 0.16 = 0.004 mol
You need moles HA after 8ml of 0.21 M KOH added = 0.004 - (0.008 x 0.21) = 0.00232 mol HA
moles of salt = moles of acid reacted = (0.008 x 0.21) = 0.00168 mol A-
Now work out concentrations based on total volume = 0.025 + 0.008 = 0.033 dm3
[HA] = 0.00232/0.033 = 0.0703 M
[A-] = 0.00168/0.033 = 0.0509 M
ka = [H+][A-]/[HA]
1.74 x 10-5 = [H+] * 0.0509/0.0703
[H+] = 2.4 x 10-5
pH = 4.62
<span>For the 40 ml addition there is no acid left over, so it's just calculated from excess base.
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