Titration calculations (1 Viewer)

[Aqua10]

Could anyone solve the following problems? Relevant working would be appreciated.


Q1. 5.24g of anhydrous sodium carbonate was dissolved in water in a volumetric
flask and the volume made up to 250 mL. 10 mL of this solution was pipetted in
a conical flask and titrated with HCl. 21.6 mL was required to reach the endpoint
Calculate the concentration of the HCl solution.

This solution was then used to determine the concentration of an unknown
barium hydroxide solution. 25 mL of of the barium hydroxide required 28.4 mL
HCl for neutralisation. Calculate the concentration of barium hydroxide solution.

Q2. In order to determine the concentration of HCl solution, a student diluted 10 mL
HCl to 250 mL then titrated 10 mL of the diluted solution with 0.147 mol/L
sodium hydroxide solution. 21.5 mL was needed to reach the end point. Calculate
the concentration of HCl.

 

[airie]

Q1. Equation of the first reaction: Na₂CO_{3 (aq)} + 2HCL_(aq) --> 2NaCl_(aq) + H₂O_(l) + CO_{2 (g)}

The number of moles of Na₂CO₃ in the solution is 5.24/(22.99+12.01+3*16), thus its concentration is [5.24/(22.99+12.01+3*16)]/0.25 = 0.25253... mol/L. So in the 10mL used in titration, there are [5.24/(22.99+12.01+3*16)]/0.25 * 0.01 = 0.002525... moles, and this is half the number of moles of HCl used, fromthe equation above. So the number of moles of HCl used is 2*0.002525... = 0.00505... moles, therefore the concentration of HCl is 0.00505... / 0.0216 = 0.2338... mol/L, approximately 0.23 mol/L (2 d.p.).

Equation of second reaction: 2HCl_(aq) + Ba(OH)_{2 (aq)} --> BaCl_{2 (aq)} + 2H₂O_(l)

From the first part, the concentration of HCl is 0.2338... mol/L, therefore there are 0.2338... * 0.0284 = 0.00664... moles in the amount of solution used. From the equation, this is double that of barium hydroxide, so there are 0.00664... / 2 = 0.00332... moles of barium hydroxide in the 25mL solution. Thus, its concentration is 0.00332... / 0.025 = 0.1328... mol/L, approximately 0.13 mol/L (2 d.p.).

Q2. Equation: HCl_(aq) + Na(OH)_(aq) --> NaCl_(aq) + H₂O_(l)

Same principle as Q1, the number of moles of Na(OH) is 0.147*0.0215 = 0.0031605 moles. This is the same as that of HCl used, so the concentration of the diluted solution is 0.0031605/0.01 = 0.31605 mol/L (0.316 mol/L to 3 d.p.), and that of the original 10mL solution is 0.31605*0.25/0.01 = 7.90125 mol/L (7.901 mol/L to 3 d.p.).

 

[Aqua10]

Thanks for the detailed work. However, there is a mistake in the calculation of molar mass of sodium carbonate. It should be 23*2 since there are two Na in each Na2C03 molecule. As a result, the answer for the first part of Q1 is 0.181 mol/L.

 

[airie]

Thanks for the detailed work. However, there is a mistake in the calculation of molar mass of sodium carbonate. It should be 23*2 since there are two Na in each Na2C03 molecule. As a result, the answer for the first part of Q1 is 0.181 mol/L.

Oops, thanks for the correction.

See, that's what happens when you try to write in vB code, even if you've got stuff written in the last line you get it wrong in the next XP

 

[ravdawg]

Thanks for the detailed work. However, there is a mistake in the calculation of molar mass of sodium carbonate. It should be 23*2 since there are two Na in each Na2C03 molecule. As a result, the answer for the first part of Q1 is 0.181 mol/L.

yeah thats what i got as well, almost, are we meant to round the asnwers we get in each part, ie if we found the molarity of the base are we meant to round it, or do we keep it in exact form. I kept it in exact form and got 0.183.
If we need to round, to how many decimals is acceptable??

 

[livOantO]

i prefer to keep it at exact form, because the roundin off the decimals for the final answer will be completely different.

-antO_

 

[xiao1985]

in science journals, you need to keep it to the smallest sig fig provided in your question (or in your observations in real life)