Author Topic: Back titration! (Read 851 times) Tweet Share

cindyy · « **on:** January 24, 2010, 09:21:39 am »

A sample of solid sodium sulfite (Na₂SO₃ $.$ XH₂O)of mass 0.4322g was dissolved in water and oxidised to sodium sulfate by adding exactly 0.8000g of I₂.

I₂ + SO₃^2- + H₂O --> 2I^- + SO₄^2- +2H⁺

The resulting solution was then neutralized by the adition of exactly 40.00mL of 0.100M NaOH. Calculate the value of X.

fady_22 · « **Reply #1 on:** January 24, 2010, 11:09:04 am »

$n(NaOH)=cV=.1*0.04=0.004 mol$
$n(H+)=n(NaOH)=0.004 mol$

So, $n(SO_{3}^{2-})=\frac{0.004}{2}=0.002 mol$ (from the equation)
$m_{r}=\frac{m}{n}=\frac{0.4322}{0.002}=216.1 gmol^{-1}$
To find the molar mass of water alone, subtract the molar mass of $Na_{2}SO_{3}$ which becomes 90 g mol-1.
Divide this by 18 g mol-1 to find the amount of H20, which is 5.

Therefore, X=5.

cindyy · « **Reply #2 on:** January 24, 2010, 03:19:40 pm »

thankkk you, i just got the answer myself

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Author Topic: Back titration! (Read 851 times) Tweet Share

cindyy

Back titration!

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