Titration - solving for a one concentration knowing the other reactant

[daewoo]

Homework Statement

A 10.02 mL solution of chemical A4+ is determined by precipitation titration with 40.07 mL of 0.03015 M titrant B3-. The reaction stoichiometry is given by

3A4+ + 4B3- => A3B4 (s)
For a log Ksp of -36.6, what is pB3- at the equivalence point?
You must express your answer to two decimal places only in order to get the program to grade it properly!(Caution: Note the difference between Log X and pX, in terms of sign)

Homework Equations

-log[concentation]

The Attempt at a Solution

since its at equivalence point the concentration of [b3]and [a] would be the same.
since ksp=[a]=-36.6
In terms of x, wouldn't it be (3x^3)(4x^4) = -36.6? (6912x^7)=-36.6 = -1.1726 but then i can't take a -log of a negative number?

 

[daewoo]

what i tried doing now was

since it said For a log Ksp of -36.6
pKsp = -log[x]
log[ksp] = -36.6
2.51x10^-37?

[3x]^3[4x]^4 = ksp
6912x^7=2.51x10^-37
x=1.67x10^-6

-log 1.67x10^-6
ph = 5.77

but that's wrong too, the answer is 5.11

so anyone have any ideas?

 

[Hunt_]

since it said For a log Ksp of -36.6
pKsp = -log[x]
log[ksp] = -36.6
2.51x10^-37?

[3x]^3[4x]^4 = ksp
6912x^7=2.51x10^-37
x=1.67x10^-6

correct

-log 1.67x10^-6
ph = 5.77

but that's wrong too, the answer is 5.11

so anyone have any ideas?

Ofcourse it is wrong. The concentration of B^{4-} at equilibrium is [4x] and not x