Titration with phenolphthalein adds 1 H^+ to the CO3^2- to make HCO3^-. You are right with mols K2CO3. That titrates none of the HCO3^-
The methyl orange end point adds another H^+ to the HCO3^- formed from the CO3^2- in the first end point. It adds a H^+ to the HCO3^- that was there initially but was not titrated in the phenolphthalein end point.
35.00 mL = all of the HCO3^- + all CO3^2-
10 mL for CO3^2- to HCO3^- + another 10 mL for that HCO3^- to H2CO3.
Therefore, 35 - 20 = 15 mL for the HCO3^- by itself.
FB1 is a solution containing a mixture of potassium carbonate and sodium hydrogen carbonate . 25cm^3 of this mixture was to titrated against 0.250mol/dm^3 solution of HCL. In one titration using phenolphthalein as indicator, 10.00cm^3 of the acid was needed. In another titration using methyl orange as indicator, 35.00cm^3 of acid was required.
Find the number of moles of potassium carbonate and sodium hydrogen carbonate present in the mixture.
I found the number of moles of potassium carbonate.
No of moles of HCL that reacted with potassium carbonate=No of moles of potassium carbonate = 0.01 x 0.250mol =0.00250mol
im not sure how to find moles of sodium hydrogencarbonate
DrBob222
answered8 years ago
Explain Bot
answered11 months ago
To find the number of moles of sodium hydrogen carbonate (NaHCO3) present in the mixture, you can use the titration data obtained using methyl orange as the indicator.
In the titration using methyl orange, it took 35.00 cm^3 of the HCl solution to completely react with the mixture. This volume represents the total amount of HCl needed to neutralize all the sodium hydrogen carbonate in the mixture.
We can calculate the number of moles of HCl used in the titration using the formula:
Moles of HCl = Volume of HCl (in dm^3) x Concentration of HCl (in mol/dm^3)
Given the volume of HCl used is 35.00 cm^3, we need to convert it to dm^3 by dividing by 1000:
Volume of HCl = 35.00 cm^3 / 1000 = 0.035 dm^3
The concentration of HCl is given as 0.250 mol/dm^3.
Plugging these values into the formula, we get:
Moles of HCl = 0.035 dm^3 x 0.250 mol/dm^3 = 0.00875 mol
Since sodium hydrogen carbonate (NaHCO3) reacts in a 1:1 ratio with HCl, the number of moles of NaHCO3 present in the mixture is also 0.00875 mol.
Therefore, the number of moles of potassium carbonate (K2CO3) present in the mixture is 0.00250 mol, and the number of moles of sodium hydrogen carbonate (NaHCO3) present in the mixture is 0.00875 mol.
In the titration using methyl orange, it took 35.00 cm^3 of the HCl solution to completely react with the mixture. This volume represents the total amount of HCl needed to neutralize all the sodium hydrogen carbonate in the mixture.
We can calculate the number of moles of HCl used in the titration using the formula:
Moles of HCl = Volume of HCl (in dm^3) x Concentration of HCl (in mol/dm^3)
Given the volume of HCl used is 35.00 cm^3, we need to convert it to dm^3 by dividing by 1000:
Volume of HCl = 35.00 cm^3 / 1000 = 0.035 dm^3
The concentration of HCl is given as 0.250 mol/dm^3.
Plugging these values into the formula, we get:
Moles of HCl = 0.035 dm^3 x 0.250 mol/dm^3 = 0.00875 mol
Since sodium hydrogen carbonate (NaHCO3) reacts in a 1:1 ratio with HCl, the number of moles of NaHCO3 present in the mixture is also 0.00875 mol.
Therefore, the number of moles of potassium carbonate (K2CO3) present in the mixture is 0.00250 mol, and the number of moles of sodium hydrogen carbonate (NaHCO3) present in the mixture is 0.00875 mol.
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