a solution of HCL is approximately 0.1M and has to be standardized . 1.234g Na2CO3 are dissolved to make 100ml of solution. 20mL of this solution are titrated with the HCL solution, requiring 31.3mL for the second equivalence point. what is the exact concentration of the HCL solution?
Mw of Na2CO3 = 105.94,
Thus moles Na2CO3 = 1.234/105.94 = 0.011.
0.011*0.1 L = 0.0011mol/L
from here on I'm confused as to whether we use M1V1=M2V2 or another method. Would appreciate some help, Thanks.
w of Na2CO3 = 105.94,
Thus moles Na2CO3 = 1.234/105.94 = 0.011.
My calculator gives 0.011648. Why did you throw those other numbers away? You have 4 significant figures in 1.234 so you should be able to have AT LEAST 4 places and I like to carry one extra place--then round a the end.
0.011*0.1 L = 0.0011mol/L
If you multiply mols x L you CAN'T get mols/L. Anyway, I would go another route
This is what I would do.
You have 0.011648 mols. Since you put it in 100 mL and used only 20 mL of it, the amount used in the titration was
0.011648 x (20/100) = ? mols and you can do the math.
The titration is
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
Convert mols Na2CO3 to mols HCl using the coefficients in the balanced equation. Then M HCl = mols HCl/L HCl. Finally round to the appropriate number of significant figures. The number is dictated by the 31.3 mL (for 3 places) unless that really is 31.30 and you just omitted the final zero. In that case round the final answer to 4 places.
Hope this helps.
Thanks for the time and help, this was my final working out, don't know if you will receive this, so once again thank you.
Step 1: moles Na2CO3 = 1.234g/105.89 g/mol = 0.0116 mol.
Step 2: 20mL of Na2CO3 = 0.0116*(20/100) = 0.00232mol/L
Step 3: Balanced equation =
Na2CO3 + 2HCL --> 2NaCl + H2O + CO2
Thus it is a 2:1 ratio, therefore moles HCl = 2* moles Na2CO3 = 2*0.00232 = 0.00464 moles.
Step 4: Conc. of HCl = 0.00464mol/ 0.03130L = 0.1469mol/L
To find the exact concentration of the HCl solution, we can use the concept of stoichiometry and the equation M1V1 = M2V2 where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.
In this case, we know the following information:
- The HCl solution has an approximate concentration of 0.1 M.
- 1.234 g of Na2CO3 is dissolved in 100 mL of solution.
- 20 mL of this Na2CO3 solution requires 31.3 mL of the HCl solution for the second equivalence point.
First, let's calculate the number of moles of Na2CO3:
Molar mass of Na2CO3 = 2 * molar mass of Na + molar mass of C + 3 * molar mass of O
= 2 * (22.99 g/mol) + 12.01 g/mol + 3 * 16.00 g/mol
= 105.94 g/mol
Moles of Na2CO3 = Mass of Na2CO3 / Molar mass of Na2CO3
= 1.234 g / 105.94 g/mol
= 0.011 mol
Since the volume taken for titration is 20 mL, which is equivalent to 0.020 L, we have:
Moles of HCl = Moles of Na2CO3
= 0.011 mol
Finally, we need to find the concentration of the HCl solution:
M2 = Moles of HCl / V2
= 0.011 mol / 0.0313 L
≈ 0.351 M
Therefore, the exact concentration of the HCl solution is approximately 0.351 M.