Data processing :
Gross volume of HCl used :
20.7 + 20.6 + 20.2 = 20.5 cm³ + 0.05 cm³.
3
Uncertainty : | 20.5-20.7 | + | 20.5-20.6 | + | 20.5-20.2 | x 100 = 2.9%
20.5
Gross volume of HCl used with percentage of uncertainty = 20.50 cm³ + 2.9% cm³.
The molarities of HCl after dilution :
M1V1 = M2V2
(2mL) (11M) = (X) (250mL)
X = 0.088 M
Uncertainty :
V1 = 2.0 mL + 0.5 mL V2 = 250.0 mL + 0.5 mL
= 0.5 x 100% = 0.5 x 100%
2.0 250.0
= 25 % = 0.2 %
Then, add together to form general uncertainty = 25 % + 0.2 %
= 25.2 %.
So the molarity of HCl after dilution is 0.088 M + 25.2 % M.
Chemical equation : 2HCl + Na2CO3 → 2NaCl + H2O + CO2
The number of moles for Na2CO3 : 1.329 = 0.0125 moles.
106
2moles of HCl+1mole of Na2CO3 → 2moles of NaCl+1mole of water+1mole of CO2
0.025mole of HCl+0.0125mole of Na2CO3 → 0.025mole of NaCl+0.0125mole of water+0.0125mole of CO2
So, the number of moles for HCl taken is 0.0250 moles.
The molarities of Na2CO3 : no. of moles
Volume
0.0125 = 0.05 M.
0.25
In this reaction, limiting reagent = Na2CO3
Theoretical moles of Na2CO3 solution :
Moles = Volume x Molarities
= (25cm³) (0.088M)
1000
= 0.0022 mole.
Actual moles of Na2CO3 used when titration :
Moles = 0.088
2
= 0.044 mole.
So the theoretical volume of HCl needed to react with 0.025 mole of Na2CO3 :
MV of Na2CO3 = (0.05) (25)
= 1.25 (2M)
= 2.5 .
1000
= 28.41 cm³.
Percentage error :
28.41cm³-20.5cm³ x 100% = 27.8423%
28.41cm³
Conclusion : The actual volume of HCl needed to neutralize the Na2CO3 is 28.41cm³ for the titration.
