Hi I have this problem I have no idea where or what to do.
25.00 mL of 0.0500 M Na2CO3 was titrated with HCl(aq) using phenolphthalein indicator, the 'end-point' (color change from pink to colorless) was reached when 10.50 mL of the acid was added. When methyl orange was used, the end-point (pink to orange) was reached when 21.00 mL of the acid was added. Account for the difference in titer volumes, and calculate the molarity of the acid.
There IS no difference in the volumes. It takes 10.50 mL of the acid to add the first H as follows:
Na2CO3 + HCl ==> NaHCO3 + NaCl
Then it takes another 10.50 mL (21.00-10.50 = 10.50 mL) to add the second hydrogen.
NaHCO3 + HCl ==> NaCl + H2CO3 (=>H2O + CO2).
(The problem did NOT say that ANOTHER 21.00 mL was added for a total volume of 21.00 + 10.50 = 31.50. If that is the case, that's a completely different problem.)
[Another note: The question about the titer could mean that you are to explain why it took twice as much for the second end point as the first end point. Of course the answer is that the first H is being added with the first 10.50 mL and the second H is being added in the next 10.50 mL and it SHOULD take twice as much provided there is no NaHCO3 impurity in the Na2CO3 sample).
As for the M of the acid,
M x L = moles.
You know how many moles were in the Na2CO3 (M x L) and you use moles Na2CO3 = M x L to calculate M HCl.
1.53molar
To account for the difference in titer volumes and calculate the molarity of the acid, we can use the concept of stoichiometry in a titration.
In a titration, a known volume and concentration of a solution (the analyte) reacts with a solution of another reactant (the titrant), where the concentration of the titrant is usually unknown. The goal of the titration is to determine the unknown concentration of the titrant.
In this case, we have Na2CO3 (sodium carbonate) as the analyte and HCl (hydrochloric acid) as the titrant. We can write the balanced chemical equation for the reaction between Na2CO3 and HCl as follows:
Na2CO3 + 2HCl -> 2NaCl + H2O + CO2
Let's analyze the titration using phenolphthalein indicator first:
- At the "end-point" (color change from pink to colorless) when using phenolphthalein, it means that all the Na2CO3 has reacted completely with HCl.
- From the balanced chemical equation, we can see that the mole ratio between Na2CO3 and HCl is 1:2. This means that for every mole of Na2CO3, 2 moles of HCl are required to react with it.
- The moles of Na2CO3 can be calculated by multiplying the volume (in liters) of the Na2CO3 solution used in the titration (25.00 mL = 0.02500 L) by the molarity (0.0500 M) of the solution:
moles of Na2CO3 = volume (L) x molarity = 0.02500 L x 0.0500 M = 0.00125 moles
- Since we have a 1:2 mole ratio between Na2CO3 and HCl, the moles of HCl required to react completely with the Na2CO3 is twice as much:
moles of HCl (phenolphthalein) = 2 x moles of Na2CO3 = 2 x 0.00125 moles = 0.00250 moles
Now, let's analyze the titration using methyl orange indicator:
- At the "end-point" (color change from pink to orange) when using methyl orange, it means that all the Na2CO3 has reacted with the excess HCl. Methyl orange is used because its color change occurs after phenolphthalein, indicating the completion of the reaction.
- From the balanced chemical equation, we know that the only difference in the reaction is the production of CO2. The moles of CO2 produced would be half of the moles of HCl used in the reaction.
moles of CO2 = 0.00250 moles (moles of HCl phenolphthalein) / 2 = 0.00125 moles
To calculate the moles of HCl (methyl orange):
- The moles of HCl required to react with Na2CO3 in excess would be equal to the moles of CO2 produced, which is 0.00125 moles.
- However, we need to account for the additional HCl volume used beyond the phenolphthalein endpoint. So, we subtract the moles of HCl required for phenolphthalein from the total moles of HCl:
moles of HCl (methyl orange) = moles of CO2 + moles of HCl (phenolphthalein)
= 0.00125 moles + 0.00250 moles
= 0.00375 moles
We can now calculate the molarity of the HCl solution:
- The volume of the HCl solution used in the titration is given as 10.50 mL = 0.01050 L.
- Using the moles of HCl (methyl orange) calculated above:
molarity of HCl = moles / volume (L)
= 0.00375 moles / 0.01050 L
= 0.357 M
Therefore, the molarity of the HCl(aq) solution is 0.357 M.