Question

DATA

Data
	Trial 1	Trial 2	Trial 3
Volume of acid (mL)	10.0	10.0	10.0
Molarity of NaOH (M)	0.20	0.20	0.20
Start NaOH buret reading (mL)	0.00	27.02	53.61
End NaOH buret reading (mL)	27.02	53.61	80.86
Volume of NaOH used (mL)	27.02	26.59	27.25

Analysis

1. Calculate the moles of NaOH used in each trial

Trial 1

Trial 2

Trial 3

2. Write a balance equation for the reaction.

3. Calculate the moles of HCl used in each trial

Trial 1

Trial 2

Trial 3

4. Calculate the molarity of the HCl

Trial 1

Trial 2

Trial 3

Answer 1

Trial 1

Volume of NaOH = 27.02 ml

Concentration of NaOH = 0.2 M

Moles of NaOH = 27.02 x 0.2 / 1000 =  0.005404 Moles

the moles of HCl used =   0.005404 Moles

According to the balanced reaction, equivalent moles of NaOH and HCl required

Volume of acid  = 10 ml  

Concentration of HCl =   0.005404 x 1000 / 10 = 0.5404 M

Trial 2

Volume of NaOH = 26.59 ml

Concentration of NaOH = 0.2 M

Moles of NaOH = 27.02 x 0.2 / 1000 =  0.005318 Moles

the moles of HCl used =   0.005318 Moles

According to the balanced reaction, equivalent moles of NaOH and HCl required

Volume of acid  = 10 ml  

Concentration of HCl =   0.005318 x 1000 / 10 = 0.5318 M

Trial 2

Volume of NaOH = 27.25 ml

Concentration of NaOH = 0.2 M

Moles of NaOH = 27.25 x 0.2 / 1000 =  0.00545 Moles

the moles of HCl used =   0.00545 Moles ​​​​​​​

According to the balanced reaction, equivalent moles of NaOH and HCl required

Volume of acid  = 10 ml  

Concentration of HCl =   0.00545 x 1000 / 10 = 0.545 M