TBR Gen Chem Titrations Q33

[ThinkDoc]

The passage gives the titration curves for a conjugate pair. The question then asks you to compare the pH at the starting point for the acid with the pH at the equivalence point for the conjugate base.

The initial concentrations of the acid and the conjugate base are .10 M in 25 ml, and you bring the volume to 50 ml by adding 25 of the titrant.

The answer key says that concentrations of the two point differs by a factor of 2, therefore the pH should differ by the log of the square root of 2. How come?

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[LoLCareerGoals]

Can you post a full question for those of us without this book? It is probably due to the coefs of the reaction (2 proton acid), hard to tell without actual reaction.

 

[Tarasaurusrex]

M1V1 = M2V2

At equivalence point

M1 = 0.1M
V1 = 0.025L
V2 = 0.05L

So M2 = 0.05M

x, -x, -x

-x^2 = (M1/M2) = -2

-x = (2)^(1/2)

x = log(2^(1/2))

 

[V5RED]

It is based on the quick formula they use in the BR books for determining pH at equivalence points.

For the -log[HA]/2 part, your [HA] for the less concentrated form is half as much as for the concentrated form. This means a difference of -log[2]/2 (ie one concentration is double of the other) this is equivalent to -log(2^1/2) because you can convert a scalar multiplied by a logarithm into a logarithm using said scalar as an exponent.

Basically they got really fancy with their answer and didn't bother to explain how they got there.