Question

I'm attempting to perform some calculations on a backwards titration. After HCl has completeley reacted with all CO₃^2- in solution, we have excess HCl which we then titrate with NaOH to determine how much carbonate was in solution to begin with. Th process was simulated by adding 2 mL of 3 M HCl to a flask and titration with NaOH. We are to assume all HCl added reacted with carbonate and are to subtract the number of mols H+ consumed by NaOH by the total amount added to obtian the simulated amount of carbonate we had in solution.

I'm a bit stuck. Here is my math so far, and I have no idea if I am even doing this right.

36.6 mL NaOH added during titration to 2 mL of 3 M HCl

$\frac{3 mol HCl}{1 L}\cdot \frac{0.0386 L}{1}\cdot \frac{1 mol NaOH}{1 mol HCl} = 0.1158 mol NaOH$

This is where I draw a blank because I realized that all I did was find the moles of OH- after adding NaOH to 3 M HCl

Answer 1

Well... I will first show you that what reaction is going on in this experiment.

NaOH + HCl = NaCl + H2O

Na2CO3 + HCl = NAHCO3+ NaCl

NAHCO3 + HCl = NaCl + CO2+ H2O

Let consider 25 ml stock solution contain full NaOH and half half NAHCO3 and it can be titrated with standard HCl solution using phenolphthalein indicator ( at the end point pH= 8.3) .The remaining half is then titrated using methyl orange indicator ( at the end point pH= 3-4 ). Now the titration is processed and shipped to the calculation process.

Here I have calculated based on normality..you could process it in molarity also. Now in your experiment you have added extra 36.6 ml HCl from outside . Where I have taken a 25 ml mixture contain full NaOH and half Na2CO3 . In your case you have to add this to the volume of NaOH .

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