[themann]

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.

[cathy]

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 ^ ^"

[cuongt]

yep ratio is 1:2 so u would need 50mL of HCl for every 25mL of Na2CO3.  

You are both rite.  

[AWK]

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.

[Borek]

See titration curve. There are two endpoints, all depends on the indicator used (although first is way to flat for a good precision).

[themann]

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?