In: Chemistry

# A beaker containing 50.0 ml of 0.300M Ca2+ at ph 9 is titrated with 0.150M EDTA....

A beaker containing 50.0 ml of 0.300M Ca2+ at ph 9 is titrated with 0.150M EDTA. Calculate the pCa2+ at the equivalence point

## Solutions

##### Expert Solution

Consider a complex formation reaction: Ca 2+ + EDTA [ CaY ] 2-

For above reaction, conditional formation constant is K' f = ( Y 4-) K f

Therefore, K' f = 0.041 x 10 10.65 = 1.83 x 10 9

First , we Calculate  of volume of EDTA required to reach equivalence point

From above reaction , we have 1 mol Ca 2+ 1 mol  EDTA

Therefore, M ca 2+ V Ca2+ = M EDTA V EDTA V EDTA = M ca 2+ V Ca2+ / M EDTA =

V EDTA = 0.300 M 50.0 ml / 0.150 M

V EDTA = 100 ml

volume of EDTA required to reach equivalence point = 100 ml

At the equivalence point, all the calcium ions are consumed by added EDTA. Hence, at this stage concentration of Ca 2+ will be due to dissociation of [CaY] 2-.

Moles of [ CaY ] 2- = moles of EDTA added to the solution

No. of moles of EDTA added in the solution = 0.150 mol / L 0.100 L = 0.015 mol

Volume of solution at equivalence point = Volume of Ca 2+ solution + volume of EDTA solution

Volume of solution at equivalence point = 50.0 + 100 ml =150.0 ml = 0.150 L

[CaY] 2-. = 0.015 mol / 0.150 L = 0.100 M

Let’s use ICE table.

 Concentration ( M ) Ca 2+ EDTA [CaY] 2- I 0.100 C +X +X -X E X X 0.100 - X

We have , K' f = [CaY2-] / [Ca 2+] [EDTA]

Therefore,K' f = 0.100 -X / (X)(X) = 1.83 10 9

0.100 -X / (X)(X) = 1.83 10 9 0.100  -X / X 2 = 1.83 10 9

At equilibrium X is negligible as compared 0.100. Then we can write 0.100 / x 2 = 1.83 x 10 9 X 2 = 0.100 / 1.83 10 9 =5.464 10 -11 X = 7.392 10 - 6 M [Ca 2+] = 7.932 10 - 6 M

We have relation, pCa 2+ = - log [Ca 2+] pCa 2+ = - log 7.932 10 - 6 = 5.13

ANSWER : pCa 2+ at equivalence point = 5.13

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