Question

Concentration of NaOH = 1600 x10- Molar Sodium hydroxide Nach Trial 1 Trial 2 Trial Trial 4 2.60/1.20 11.40/1.30 28.20 29.10.26.1026.50 Initial buret reading Final buret reading Volume of NaOH added Moles of NaOH added Moles of HCl reacted Molarity of diluted HCI Molarity of undiluted HCI Average molarity of undiluted HCI Precision

Answer 1

Calculations:

• 1 mol of NaOH = 1 mol of HCl
• NaOH concentration = 0.100 M
• Moles of NaOH added : volume of NaOH added (ml) * 0.100 moles / 1000 ml
• M₁V₁ = M₂V₂ :

Molarity of HCl = Molarity of NaOH *volume of NaOH added (ml) / volume of HCl added (ml)

	Trial 1	Trial 2	Trial 3	Trial 4
Volume of NaOH (ml)	25.6	27.9	24.70	25.20
Moles of NaOH	2.56*10-3	2.79*10-3	2.47*10-3	2.52*10-3
Moles of HCl	2.56*10-3	2.79*10-3	2.47*10-3	2.52*10-3
Molarity of Diluted HCl	0.1024	0.1116	0.0988	0.1008
Molarity of Un-diluted HCl	1.024	1.116	0.988	1.008
Average Molarity of Un-diluted HCl	1.034
Precision	1.034 ± 0.108 (range method)

Calculation (examples) :

# Trial 1 :

• Moles NaOH : 25.6 ml * 0.100 moles / 1000 ml = 2.56*10^-3 mol
• Molarity of Diluted HCl = 0.100 *25.6(ml) / 25 (ml) = 0.1024 M
• Molarity of Un-diluted HCl : dilution factor = 10

so, 10*0.1024 M = 1.024 M