Qualification > Sciences
IG Chemistry Paper 3 discussion
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what i think was that the value of the height had nothing to do with the volume, since height was in mm, and volume in cm3. I just remember that, 18 cm3 of aqueous nitrate of T, reacted with 2 cm3 portions of a solution of Na3PO4. The maximum precipitate of the phosphate of T, was reached at 6 cm3 of sodium phosphate. Since both had the same concentration, it didnt really matter, the main thing that mattered was the volume then. Now, we have that the volume of the nitrate of T used was 18, and that of the sodium phosphate was 6. so volume ratio of the nitrate reactant to the phosphate reactant is 18:6 = 3:1, which is also the mole ratio. So if we construct an ionic equation using the mole ratio, we can get ...
PO43- + 3T+ ----> T3PO4
when i came home, i opened june 2004 paper 3, of chem, where i remember seeing a similar graph as the one given in the question..
in the june 2004 q, 4 cm3 of aqueous iron (III) chloride, reacts with 2cm^3 portions of aqueous sodium hydroxide..according to the graph..the maximum volume of sodium hydroxide at which the maximum precipitate is formed, is at 12 cm^3..
thus the volume ratio of the chloride reactant to the hydroxide, is 4:12 = 1:3
That means, there will be one Fe3+ ion for every 3 OH- ions..according to the ratio..
the next part of the given is..complete the following reaction..
Fe3+ + .....OH- -----> ...........
since ratio is 1:3..itll be 3 OH- forms Fe(OH)2..
the concentration of both reactants is 1moldm-3 which is given..
so the questions are quite similar..and so, i think T3PO4
i dont consider myself a genius, just mixxing familiar stuff with unfamiliar stuff.
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sorry...that is Fe(OH)3
Shogun:
it wasnt OH in tht fe thing it was some halogen i think, im100% sure it wasnt OH in the fill in the blanks stuff for tht ionic reaction. however ure explaination is good.
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i just rechecked it..and it is aqueous sodium hydroxide, and aqueous iron (III) chloride..as the reactants..:S
shafiq_92libra:
At da cathod(da negativ electrod) u gt H2+ ions atraced n dey giv H2 gas at cathod. At da anode wich is positiv, da negtiv ions dat r Cl- ions r atracted n dey giv off Cl2 gas. A solution of NaOH is left behind ;D
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