Na2CO3 is determined from the second part of the titration. The equation is
NaHCO3 + HCl ==> NaCl + CO2 + H2O
mass Na2CO3 = moles Na2CO3 x molar mass Na2CO3.
moles Na2CO3 = MHCl x LHCl
Therefore, mass Na2CO3 =
MHCl x LHCl x molar mass Na2CO3 and
percent Na2CO3 is (mass Na2CO3*100)/2.0 g.
The second one is similar.
V1 titrates all the OH and 1/2 the carbonate half way (to bicarbonate).
V2 titrates the bicarbonate to carbonate (half way, that is); therefore, V1 + V2 titrates all the oH and all the carbonate.
So we have the same 100/1000 factor (100 converts to percent and 1000 converts from mL of V1+V2 to L of V1 + V2), then 40 (the molar mass NaOH) divided by 10 = 4.0.
What your prof has done is to substitute 1 M HCl for the titrant, mL for L HCl (but note mL must be divided by 1000 to convert to L), then we multiply by 100 to convert to percent. That gives us a 100/1000 factor which is 1/10 and 1/10 x molar mass Na2CO3 (which is 106) gives 10.6.
To determine the sodium carbonate content & total alkali in NaOH, the following steps were performed:-
1) 2.0g of NaOH was dissolved in 80ml of CO2 free water.
2) 3 drops of phenolphthalein indicator were added and the solution was titrated with 1M HCl. Colour changed from pink to colourless.(V1 ml of HCl used)
3) 3 drops of methyl orange indicator were then added and the titration with 1M HCl was continued. Colour changed from yellow to red.(V2 ml of HCl used)
Calculations:
(%w/w)Na2CO3
= (V2 x 10.6 x 1M) / 2.0g
(%w/w)Total alkali
= ((V1+V2)x 4.0 x 1M) / 2.0g
Please explain how the calculation steps are derived?
Thanks.
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if i spill a little of sodium carbonate after it is weighed, will it affect my the percentage of sand that i calculate at the end of the experiment
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