Calculating K+ with an Mg chelate?
The K+ in a 250 mL sample of mineral water was precipitated with NaB(C6H4)4 ^ 1-
The precipitate was filtered, washed, and redissolved in an organic solvent. An excess of the mercery(II)/EDTA chelate was added.
The liberated EDTA was titrated with 29.64 mL of 0.05581 M Mg2+. Calculate the K+ ion concentration in ppm.
WORK I'VE DONE:
Okay I figured out the balanced equations to be:
K+ + B(C6H4)4 ^(-1) -----> KB(C6H4) (s)
and
4HgY^(2-) + B(C6H4)4 ^(-1) + 4H2O ------> H3BO3 + 4C6H5Hg^(1+) + 4HY^ (3-) + OH^(1-)
I know those are right, but I don't know what to do with them.
Answer:
Well, provided your work is correct, this is how you'd proceede:
Your second reaction shows that each mole of B(C6H4)4^(-1) releases 4 moles of EDTA (as HY^(3-)).
Adding your first reaction, the net result is per each mole of K+ you get 4 moles of EDTA.
EDTA chelates Mg(II) cations in a 1:1 ratio, the same way Hg(II) cations are chelated.
So:
0.02964 L of a 0.05581 M solution of Mg(II) contain 1.654 mmoles of Mg(II).
This means 1.654 mmoles of EDTA were released by the adition of KB(C6H4).
And this means 1.654/4 = 0.4135 mmoles of KB(C6H4) were added, containing a total of 0.4135 mmoles of K+.
ppm is equivalent to mg/L. Potassium's AW is 39.10 g/mole (= mg/mmole), so 0.4135 mmoles of K+ weight 0.4135*39.10 = 16.17 mg. And since we assayed a 0.25 L sample, the concentration of K+ is 16.17/0.25 = 64.68 ppm.
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The precipitate was filtered, washed, and redissolved in an organic solvent. An excess of the mercery(II)/EDTA chelate was added.
The liberated EDTA was titrated with 29.64 mL of 0.05581 M Mg2+. Calculate the K+ ion concentration in ppm.
WORK I'VE DONE:
Okay I figured out the balanced equations to be:
K+ + B(C6H4)4 ^(-1) -----> KB(C6H4) (s)
and
4HgY^(2-) + B(C6H4)4 ^(-1) + 4H2O ------> H3BO3 + 4C6H5Hg^(1+) + 4HY^ (3-) + OH^(1-)
I know those are right, but I don't know what to do with them.
Answer:
Well, provided your work is correct, this is how you'd proceede:
Your second reaction shows that each mole of B(C6H4)4^(-1) releases 4 moles of EDTA (as HY^(3-)).
Adding your first reaction, the net result is per each mole of K+ you get 4 moles of EDTA.
EDTA chelates Mg(II) cations in a 1:1 ratio, the same way Hg(II) cations are chelated.
So:
0.02964 L of a 0.05581 M solution of Mg(II) contain 1.654 mmoles of Mg(II).
This means 1.654 mmoles of EDTA were released by the adition of KB(C6H4).
And this means 1.654/4 = 0.4135 mmoles of KB(C6H4) were added, containing a total of 0.4135 mmoles of K+.
ppm is equivalent to mg/L. Potassium's AW is 39.10 g/mole (= mg/mmole), so 0.4135 mmoles of K+ weight 0.4135*39.10 = 16.17 mg. And since we assayed a 0.25 L sample, the concentration of K+ is 16.17/0.25 = 64.68 ppm.
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