Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: 1800chisaki on May 11, 2020, 08:56:57 AM
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The lab:
Weigh 0.0850 g KBrO3 and dissolve in 30 mL water in a 250 mL conical flask.
Then add 2.0 g KI and 10 mL HCl 2M
Titrate immediately with the Na2S2O3 solution with continuous shaking until a pale yellow color of the iodine is left.
Dilute with distilled water to about 100 mL.
Hereafter add starch solution (20drops = 1mL) and continue titrating until blue turns to colorless.
It is important to add starch just before the equivalence point (color of the solution is pale yellow), as it is insoluble with excess I2 complexes. After adding starch, the solution turns blue.
33mL Na2S2O3 was titrated.
Calculating f:
(M x f x V)Na2S2O3 = (m/Mr)KBrO3
f Na2S2O3 = (m/Mr)KBrO3 / (M x V)Na2S2O3
=(0.0850g/167,00g/mol)/(0,1mol/L x 0,033 L)
=0,154237
My questions:
i used 0.0850g as mass of KBrO3 but is this really correct? im doubting it since it was dissolved in 30 mL distilled water, im not sure if this changes the mass that im supposed to use for my calculation
One more question, the stoichiometry is
BrO3- + 9 I- + 6 H+ -> 3 I3- + Br- + 3 H2O
I3- + 2 S2O3^2- -> 3 I- + S4O6^2-
Do i need to multiply (m/Mr)KBrO3 by 9?
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What should change the mass? The 0,085 g will not change.
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What should change the mass? The 0,085 g will not change.
One more question, the stoichiometry is
BrO3- + 9 I- + 6 H+ -> 3 I3- + Br- + 3 H2O
I3- + 2 S2O3^2- -> 3 I- + S4O6^2-
Do i need to multiply (m/Mr)KBrO3 by 9?
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1 bromate gives 3 triiodide,
1 triiodide needs 2 thiosulphate
In sum you need 3 times thiosulfate what gives the faktor 6
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1 bromate gives 3 triiodide,
1 triiodide needs 2 thiosulphate
In sum you need 3 times thiosulfate what gives the faktor 6
Oh then it needs to be multiplied by 6
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Correct.
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Correct.
Thank you!