Question

"A solution that possibly contains NaOH, Na2CO3, or NaHCO3 either alone or in a compatible combination is analyzed by titration with 0.1202 M HCl.

The first titration uses 25.00 mL of the sample solution and requires 15.67 mL of HCl to reach a phenolphthalein end-point.

The second titration uses another 25.00 mL portion of the sample and requires 42.13 mL of HCl to reach a bromocresol green end-point.

Calculate the number of milligrams of each solute per milliliter of solution. "

How do I know which endpoint is neutralizing what specie. What is the importance of the volume of phenolphthalein and the volume of bromocresol green?

Answer 1

Using phenolphthalein we get end point for NaOH and HCl titration

First 25 ml sample

moles HCl added = 0.1202 M x 15.67 ml = 1.8835 mmol

mg of NaOH present = 1.8835 mmol x 40 g/mol = 75.34 mg

NaOH (mg/ml) = 75.34 mg/25 ml = 3.014 mg/ml

Second 25 ml sample

Bromocresol green used for NaHCO3 end point detection

moles HCl added = 0.1202 M x 42.13 ml = 5.0640 mmol

mg of NaOH present = 5.0640 mmol x 84.01 g/mol = 425.43 mg

NaOH (mg/ml) = 425.43 mg/25 ml = 17.017 mg/ml

The remaining is Na2CO3 in solution