Volumetric Analysis Help

[user_5]

In order to standardise a solution of hydrochloric acid, a student titrated the solution against 20.00mL aliquots of a standard solution of sodium carbonate. Methyl orange was used to identify the end point of the reaction:

2HCl + Na2CO3 -> 2NaCl + H2O + CO2

The sodium carbonate solution had been prepared by dissolving 1.236g of anhydrous Na2CO3 in water making the solution up to 250.0mL
The average titre was 20.99mL

a) What is the molarity of the sodium carbonate solution?

n(Na2CO3) = 1.236/106 = 0.01166 mol
c(Na2CO3) = 0.01166/0.250 = 0.04664 M


c) Calculate the concentration of the HCl

n(HCl) = 2* n(Na2CO3) = 2*0.01166 = 0.02332 mol
c(HCl) = 0.02332/0.02099 = 1.111 M


However, that part c) is wrong, the answer is 0.8892 M
Could someone explain ?   

[luken93]

In order to standardise a solution of hydrochloric acid, a student titrated the solution against 20.00mL aliquots of a standard solution of sodium carbonate. Methyl orange was used to identify the end point of the reaction:

2HCl + Na2CO3 -> 2NaCl + H2O + CO2

The sodium carbonate solution had been prepared by dissolving 1.236g of anhydrous Na2CO3 in water making the solution up to 250.0mL
The average titre was 20.99mL

a) What is the molarity of the sodium carbonate solution?

n(Na2CO3) = 1.236/106 = 0.01166 mol
c(Na2CO3) = 0.01166/0.250 = 0.04664 M


c) Calculate the concentration of the HCl

n(HCl) = 2* n(Na2CO3) = 2*0.01166 = 0.02332 mol
c(HCl) = 0.02332/0.02099 = 1.111 M


However, that part c) is wrong, the answer is 0.8892 M
Could someone explain ?   

You are only titrating it against 20ml of the 250ml Standard Solution.

Multiply your n(Na2CO3) by 20/250 and then try again

[bawse]

In order to standardise a solution of hydrochloric acid, a student titrated the solution against 20.00mL aliquots of a standard solution of sodium carbonate. Methyl orange was used to identify the end point of the reaction:

2HCl + Na2CO3 -> 2NaCl + H2O + CO2

The sodium carbonate solution had been prepared by dissolving 1.236g of anhydrous Na2CO3 in water making the solution up to 250.0mL
The average titre was 20.99mL

a) What is the molarity of the sodium carbonate solution?

n(Na2CO3) = 1.236/106 = 0.01166 mol
c(Na2CO3) = 0.01166/0.250 = 0.04664 M


c) Calculate the concentration of the HCl

n(HCl) = 2* n(Na2CO3) = 2*0.01166 = 0.02332 mol
c(HCl) = 0.02332/0.02099 = 1.111 M


However, that part c) is wrong, the answer is 0.8892 M
Could someone explain ?   

You are only titrating it against 20ml of the 250ml Standard Solution.

Multiply your n(Na2CO3) by 20/250 and then try again

Do you mean 250/20?

[luken93]

He has 0.01166 mol in 250ml
Divide this by 250 to get mol/mL
Multiply by 20 to get mol/20mL

Alternatively, V = n/c from your molarity

[huaxiadragon]

In order to standardise a solution of hydrochloric acid, a student titrated the solution against 20.00mL aliquots of a standard solution of sodium carbonate. Methyl orange was used to identify the end point of the reaction:

2HCl + Na2CO3 -> 2NaCl + H2O + CO2

The sodium carbonate solution had been prepared by dissolving 1.236g of anhydrous Na2CO3 in water making the solution up to 250.0mL
The average titre was 20.99mL

a) What is the molarity of the sodium carbonate solution?

n(Na2CO3) = 1.236/106 = 0.01166 mol
c(Na2CO3) = 0.01166/0.250 = 0.04664 M


c) Calculate the concentration of the HCl

n(HCl) = 2* n(Na2CO3) = 2*0.01166 = 0.02332 mol
c(HCl) = 0.02332/0.02099 = 1.111 M


However, that part c) is wrong, the answer is 0.8892 M
Could someone explain ?   

You are only titrating it against 20ml of the 250ml Standard Solution.

Multiply your n(Na2CO3) by 20/250 and then try again

Luken is very right

n(Na2CO3)(titrated)=0.02L * 0.04664M= 9.3283 * 10^-4

n(HCl)=1.86566 * 10^-3 mol
M(HCl)=1.8656 * 10^-3 / 0.02099L
M(HCl)=0.08888M (very lucky number yes? AND to 4 Sig figs I believe)

The answer is close enough I think....