Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: themann on March 23, 2006, 05:16:18 AM
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Once again i am unsure. If i have 25mL of Na2CO3 0.1M solution How much HCl would i approximately need to add. Say as a ratio from 1 mole of HCL i get 1 +H ion will i get 1 -OH ion as a result or will i get 2 or more. It seems to my understanding that it would be a 1 to 1 ratio but once again i am unsure. If it was a 1 to 2 ratio it would mean that 50mL of HCL would be needed for every 25mL of Na2CO3 solution titrated.
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I think :::
Since Na2CO3+2HCl-->CO2+H2O+2NaCl
1 mole of NaCO3 reacts with 2 mole of HCl
no. of mole of Na2CO3=molarity * volume(in dm3)
=25*10-3mole
So that no. of mole of HCl= no. of mole of Na2CO3 * 2
=5*10-3
volume of HCl used = no. of mole of HCl / molarity of HCl
if you used 0.1M of HCl,
then volume of HCl = 50mL
But I am not sure is this correct ^ ^"
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yep ratio is 1:2 so u would need 50mL of HCl for every 25mL of Na2CO3.
You are both rite. ;)
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Na2CO3 reacts with HCL in two steps
Na2CO3 = HCl = NaHCO3 + NaCl
NaHCO3 + HCl = H2O + CO2(g) + NaCl
For each step you need equimolar amount of HCl
If you use 0.1 M HCl then you need twice 25 ml od acid for 25 ml of 0.1 M Na2CO3.
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See titration curve. There are two endpoints, all depends on the indicator used (although first is way to flat for a good precision).
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ok great thanks guys. Beacause i am given 2 indicators methyl orange and phenylphthaline. The two endpoints show that the first would be detected by phenylphthaline and the second with methyl orange. so am i best off just using methyl orange to get a better precision. Does the CO2 pose a problem to the H20 which in essence might form HCO3? or something of the eqivalent?