A0.210 mol dm-3 solution of potassium hydroxide was added from a burette to 25.0cm3 of a 0.160 mol dm-3 solution of ethanoic acid in a conical flask. given that the value of the acid dissociation constant for ethanoic acid is 1.74x10-5 mol dm-3, calculate the ph at 25oc after 8cm3 of potassium hydroxide solution was added

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

For the 40 ml addition there is no acid left over, so it's just calculated from excess base.



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Where is the following?

I'm pretty sure the answer is alcohol