CHM 261- Batzer
Chem Final Write Up- Kool Aid Lab
Total Acid
This part of the lab requires titration of Kool Aid with an NaOH solution. The reason titrations are used is to determine the concentration of an unknown solution by reacting a strong acid with a strong base. Titrations are hard to accomplish, though, due to the fact that indicators used to show the endpoints are very sensitive and one drop could make the solution titrate past its endpoint.
Blue stock solution: 3 mg/L
Yellow stock solution: 35 mg/L
Phenolphthalein is the indicator used (6 drops per titration)
Molarity of NaOH used: .05056 M
Start with 25 mL green Kool Aid
Dillution factor: 100mL/25mL green Kool Aid= 4
Titration 1: Start= 0.00 mL End= 14.75 mL 14.75 mL …show more content…
Error
((14.80 mL-14.55 mL)/(14.7 mL)) x100 = 1.701%
Hydrogen Ion: (Molarity NaOH) x (Volume NaOH) = mol NaOH [1:1 ratio between NaOH & H+]= mol H+
Titration 1: (.05056 mol/L) x (.01475 L)= .00074576 mol NaOH = .00074576 mol H+
Titration 2: (.05056 mol/L) x (.01455 L)= .00073565 mol NaOH = .00073565 mol H+
Titration 3: (.05056 mol/L) x (.01480 L)= .00074829 mol NaOH = .00074829 mol H+
Hydrogen Ion Per Liter: (Mols/L of H+ 25 mL = .025 L x 40 = 1L)
Titration 1: (.00074576 mol H+/ .025 L) x 40 = .0298304 mol H+/L
Titration 2: (.00073565 mol H+/ .025 L) x 40 = .029426 mol