1. Consider a 1.00g unknown sample that contains 62.0% sodium carbonate and 38.0% sodium bicarbonate, NaHCO3, by mass. When this sample is dissolved in water, how many mL of 0.200M HCl will be required to convert all of the carbonate to bicarbonate? How many mL will be required to convert the resulting bicarbonate solution to carbonic acid, H2CO3(aq)?
2. Based on your answer to question 1, what feature or features of your titration curve will tell you whether you have a single base or a mixture of bases?
I have done titrations before but I don't understand what to do with the percentages and what they really mean. As well as answering the question, if possible, explain what the percentages mean. Thanks!
The percentages just tell you how much Na2CO3 and NaHCO3 you start with.
1 x 0.62 =0.62g Na2CO3.
1 x 0.38 = 0.38g NaHCO3.
I will estimate from here but you confirm and use better accuracy. It makes the fractions come out better.
mols Na2CO3 = 0.62/106 = 0.00585
mols NaHCO3 = 0.38/84 = 0.00452
So we place 0.00585 mols Na2CO3 and 0.00452 mols NaHCO3 in a flask and titrate with 0.2M HCl.
Na2CO3 + HCl ==> NaHCO3 + NaCl. Note that the NaHCO3 initially there isn't titrated at all.
It will take 0.00585 mols HCl to change all of the CO3^2- to HCO3^-. M = mols/L so
L HCl = mols/M = 0.00585/0.2 = 0.0292L or 29.2 mL to titrate all of the carbonate to bicarbonate.
For part 2 of #1, we now have 0.00585 mols HCO3^- from the CO3^- titration (halfway) + 0.00452 mols HCO3^- there initially for a total of 0.0104
NaHCO3 + HCl ==> NaCl + H2O + CO2
0.0104 mol NaHCO3 will require 0.0104 mols 0.2M HCl. How many L is that?
M = mols/L or L = mols/M = 0.0104/0.2 = 0.05185 = 51.85 mL of 0.2M HCl for the second titration. So total = 29.2 from first + 51.85 for the second = about 81 mL. What message do you get from this? If the mixture were not a mixture (all Na2CO3) how much would it have taken? If it took 29.2 for the first H^+ to be added ti must take double that or 58.2 total. BUT the total is 81 so something must have been mixed with the Na2CO3 to start. These numbers will come out more exact if you use better numbers as you go through the calculation.
To answer these questions, we need to understand the composition of the unknown sample and the chemical reactions involved.
1. First, let's calculate the mass of sodium carbonate (Na2CO3) and sodium bicarbonate (NaHCO3) in the 1.00g sample:
Mass of Na2CO3 = 0.62g (62.0% of 1.00g)
Mass of NaHCO3 = 0.38g (38.0% of 1.00g)
Now, let's find the moles of Na2CO3 and NaHCO3 using their molar masses:
Molar mass of Na2CO3 = 105.99 g/mol
Molar mass of NaHCO3 = 84.01 g/mol
Moles of Na2CO3 = (mass of Na2CO3) / (molar mass of Na2CO3)
= 0.62g / 105.99 g/mol
= 0.0058 mol
Moles of NaHCO3 = (mass of NaHCO3) / (molar mass of NaHCO3)
= 0.38g / 84.01 g/mol
= 0.0045 mol
The balanced chemical equation for the reaction between HCl and Na2CO3 is:
2HCl(aq) + Na2CO3(aq) -> 2NaCl(aq) + H2O(l) + CO2(g)
According to the stoichiometry of the reaction, 2 moles of HCl are required to react with 1 mole of Na2CO3. Therefore, to convert all of the carbonate (Na2CO3) to bicarbonate (NaHCO3), we need double the amount of moles of HCl.
Moles of HCl needed = 2 * Moles of Na2CO3
= 2 * 0.0058 mol
= 0.0116 mol
Now, to calculate the volume of 0.200M HCl required, we can use the equation:
Moles = Molarity x Volume (in liters)
Volume (in liters) = Moles / Molarity
= 0.0116 mol / 0.200 mol/L
= 0.058 L
To convert the volume to milliliters, we multiply by 1000:
Volume (in mL) = 0.058 L x 1000
= 58 mL
Therefore, 58 mL of 0.200M HCl is required to convert all of the carbonate to bicarbonate.
For the second part of the question, the balanced chemical equation for the reaction between HCl and NaHCO3 is:
HCl(aq) + NaHCO3(aq) -> NaCl(aq) + H2O(l) + CO2(g)
Again, we can use the stoichiometry of the reaction to determine the moles of HCl required to convert the resulting bicarbonate solution to carbonic acid (H2CO3).
Moles of HCl needed = Moles of NaHCO3
= 0.0045 mol
Using the same equation as before, we can calculate the volume of 0.200M HCl required:
Volume (in L) = 0.0045 mol / 0.200 mol/L
= 0.0225 L
Volume (in mL) = 0.0225 L x 1000
= 22.5 mL
Therefore, 22.5 mL of 0.200M HCl is required to convert the resulting bicarbonate solution to carbonic acid.
2. The percentages given for the sodium carbonate and sodium bicarbonate in the sample represent their mass percentage in the unknown sample. In other words, it tells us the proportion of each compound in the 1.00g sample. These percentages are useful in determining the number of moles and mass of each compound.
To identify the feature or features of the titration curve that can reveal whether you have a single base or a mixture of bases, you need to analyze the shape of the curve. In a titration, the titration curve shows the change in pH with respect to the volume of the titrant (in this case, HCl) added.
If you have a single base, the titration curve will exhibit a single equivalence point, where the pH changes rapidly. The volume of the titrant at the equivalence point will correspond to the amount needed to fully react with the base.
If you have a mixture of bases, you may observe multiple equivalence points or a broader, less distinct equivalence point. This is due to the presence of different bases reacting at different rates or with different stoichiometries.
By analyzing the shape of the titration curve, you can determine whether you have a single base or a mixture of bases in your unknown sample.