[zanio]

Hi all,

my name is Ana and I am new hear. I work at chemical lab and I have a problem with quantification  of Na2CO3 in presens of NaOH in Zn/Ni electrolyte.

Could you help me?

The method who we use is:

1. 5 ml of sample is add in 250 ml conical flask. Then we put 100 ml H20 and 2-3 drops of phph as an indicator. Titration with 1N HCl. ( v1)

2. 5 ml of sample is add in 250 ml conical flask. Then we put 20 ml 10% BaCl2, 100 ml H2O and 2-3 drops phph. Titration with 1N HCl. (v2)

Calculation of results:

Na2CO3, g/l = ( V2*NHCl - V1*NHCl).21.2

NaOH, g/l = V2*NHCl * 8

When we send the sample in other lab the resulds were diferend. The amound of Na2CO3 was double.


THANK YOU!

[mjc123]

Your expression for Na₂CO₃ is wrong. It should be
Na₂CO₃, g/l = (V1*NHCl - V2*NHCl)*10.6
You seem to be forgetting that 1 mol Na₂CO₃ reacts with 2 mol HCl. So, if the molar concentrations of NaOH and Na₂CO₃ are x and y, then
v1*1 = 5*(x + 2y)
y = (v1 - 5x)/(2*5) = (v1 -5v2/5)/10
C = 106*(v1 - v2)/10