Question

A titration to standardize a NaOH titrant by titrating it against the monoprotic primary standard base “TRIS” is performed in lab, with the following data being gathered:

Titration of 25.00 ± 0.03 mL TRIS solution (0.1102± 0.0004 mol/L)
Initial buret reading: 0.42 ± 0.01 mL
Final buret reading: 15.77 ± 0.01 mL

The concentration of the NaOH titrant is: ( ) ± ( ) mol/L.

Answer 1

The Tris concentration is 0.1102 0.0004 mol/L; the volume of NaOH solution required for the titration is (15.77 – 0.42) 0.01 mL = 15.35 0.01 mL.

Millimoles of Tris required for titration = (25.00 mL)*(0.1102 mol/L) = 2.755 mmole (ignore uncertainty here).

Consider the reaction now.

Tris (aq) + NaOH (aq) --------> Na-Tris (aq) + H₂O (l)

As per the stoichiometric equation,

1 mole Tris = 1 mole NaOH

Therefore, millimoles of NaOH neutralized = millimoles of Tris added = 1.69157 mmole.

Volume of NaOH required for the titration = 15.35 0.01 mL.

Molarity of NaOH (without uncertainty) = (2.755 mmole)/(15.35 mL) = 0.17948 M ≈ 0.1795 mol/L.

Next perform a progression of error analysis.

Relative uncertainty in volume of Tris taken = ΔV₁/V₁ = (0.03 mL)/(25.00 mL) = 0.0012.

Relative uncertainty in concentration of Tris = ΔM₁/M₁ = (0.0004 mol/L)/(0.1102 mol/L) = 0.0036298.

Relative uncertainty in volume of NaOH = ΔV₂/V₂ = (0.01 mL)/(15.35 mL) = 0.00065146.

Relative uncertainty in concentration of NaOH = ΔM₂/M₂ = √(ΔV₁/V₁) + (ΔM₁/M₁) + ΔV₂/V₂) = √(0.0012)² + (0.0036298)² + (0.00065146)² = 0.0038781.

Uncertainty in the concentration of NaOH = (ΔM₂/M₂)*M₂ = (0.0038781)*(0.1795 mol/L) = 0.0006961 mol/L ≈ 0.0007 mol/L.

The concentration of NaOH is 0.1795 0.0007 mol/L (ans).