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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?

Feb 22nd, 2015

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).

So first we need to know the number of mols of HA

mols HA = 0.02500 L * 0.0800 M = 2.00E-3

At the equivalence point, all HA and OH- is used up in neutralization to make H2O

All that is left is A-. This A- will react with water to create more HA until equilibrium is reached (called hydrolysis).

Using an ICE table, we can find that the A- remaining after equilibrium will be given by

1.8E-5 = x^2 / (2E-3 - x)

using successive approximations, x=1.8095E-4

mols A- = 2E-3 - 1.8095E-4 = 1.819E-3

[A-] = 1.819E-3 / 0.02857 L = 6.37E-2 M

Feb 22nd, 2015

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Feb 22nd, 2015
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Feb 22nd, 2015
Sep 25th, 2017
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