titration of 25.00 mL of a 0.0800 M acetic acid solution with 0.0700 M KOH. assume temperature is 25 oC.
What is the analytical concentration of acetate ions at the equivalence point?
What volume of KOH has been added at the half-equivalence point?
OK, so the equivalence
point is the point on the titration curve where the mols of OH- added equals
the mols of HA (I'm using HA to refer toanygeneral weak acid).
first we need to know the number of mols of HA
= 0.02500 L * 0.0800 M = 2.00E-3
equivalence point, all HA and OH- is used up in neutralization to make H2O
that is left is A-. This A- will react with water to create more HA until
equilibrium is reached (called hydrolysis).
an ICE table, we can find that the A- remaining after equilibrium will be given
= x^2 / (2E-3 - x)
successive approximations, x=1.8095E-4
= 2E-3 - 1.8095E-4 = 1.819E-3
1.819E-3 / 0.02857 L = 6.37E-2 M
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