A .523g sample of a mix of Na2co3 and Nahco3, is titrated with .1 HCl requiring 17 ml to reach the phenolphthalein end point and a total of 43.8 ml to reach the methyl orange end point. What is the percent of each of na2co3 and nahco3 in the mixture?
Here is the way this works.
At the beginning of the titration you have a mixture of CO3^2- + HCO3^-.
When you titrated with the first 17 mL you don't touch the HCO3^-; you titrate all of CO3^- halfway. That is
CO3^2- + H^+ ==> HCO3^-. With that information you can calculate the percent Na2CO3.
mL x M x milliequivalent weight = 17 x 0.1 x 0.106 = ? grams Na2CO3.
Then %Na2CO3 = (g Na2CO3/mass sample)*100 = ?
The next part of the titration titrates what's left. What's left is all of the CO3^- (which is now HCO3^-) + the HCO3^- there from the start. It takes 43.8 mL to titrate from beginning to end of which 34 mL (17 mL x 2) = the volume needed to titrate the carbonate. (It take 17 mL to titrate the carbonate half way so it takes 34 mL to titrate it all the way). So 43.8-34 = 9.8 mL to titrate the HCO3^- that was in the sample initially.
grams HCO3^- = 9.8 x 0.1 x 0.084 = g.
%NaHCO3 = (mass NaHCO3/mass sample)*100 = ?
thank you DrBob222
To determine the percent of Na2CO3 and NaHCO3 in the mixture, we need to understand the reaction that occurs during the titration.
The balanced chemical equation between Na2CO3 and HCl is as follows:
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
Now, let's calculate the number of moles of HCl that reacted with the sample.
Number of moles of HCl = (Volume of HCl used / 1000) x Concentration of HCl
Since we know the volume of HCl used for each endpoint, we can calculate the number of moles of HCl for both endpoints.
For the phenolphthalein endpoint:
Number of moles of HCl at phenolphthalein endpoint = (17 mL / 1000) x (0.1 mol/L) = 0.0017 mol HCl
For the methyl orange endpoint:
Number of moles of HCl at methyl orange endpoint = (43.8 mL / 1000) x (0.1 mol/L) = 0.00438 mol HCl
Next, we need to determine the number of moles of Na2CO3 and NaHCO3 in the mixture based on the reaction stoichiometry. Let's assume the number of moles of Na2CO3 is "x" and the number of moles of NaHCO3 is "y."
From the balanced equation, we can see that for every 1 mole of Na2CO3, 2 moles of HCl react. Therefore, the moles of NaHCO3 can be calculated as follows:
0.0017 mol HCl = 2x
Solving this equation, we get:
x = 0.00085 mol Na2CO3
Substituting the value of x into the equation, we have:
y = 0.0017 - 2x
y = 0.0017 - 2(0.00085)
y = 0.0017 - 0.0017
y = 0
This means that there is no NaHCO3 present in the mixture.
Now, let's calculate the percent of Na2CO3 in the mixture.
Percent of Na2CO3 = (Mass of Na2CO3 / Total mass of mixture) x 100
To calculate the mass of Na2CO3, we can use the following equation:
Mass of Na2CO3 = (Number of moles of Na2CO3) x (Molar mass of Na2CO3)
The molar mass of Na2CO3 can be calculated as follows:
Molar mass of Na = 22.99 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Total molar mass of Na2CO3 = (2 x Molar mass of Na) + Molar mass of C + (3 x Molar mass of O)
= (2 x 22.99) + 12.01 + (3 x 16.00)
= 105.99 g/mol
Substituting the values, we have:
Mass of Na2CO3 = 0.00085 mol Na2CO3 x 105.99 g/mol
= 0.09065 g Na2CO3
The total mass of the mixture is given as 0.523 g.
Let's substitute the values into the equation to calculate the percent of Na2CO3:
Percent of Na2CO3 = (0.09065 g Na2CO3 / 0.523 g) x 100
= 17.34%
Therefore, the percent of Na2CO3 in the mixture is approximately 17.34%, and there is no NaHCO3 present.
To determine the percent of Na2CO3 and NaHCO3 in the mixture, we need to use the concept of stoichiometry and the volumetric analysis data provided.
Here's how we can calculate the percent of each component in the mixture:
1. Calculate the moles of HCl used to reach the phenolphthalein and methyl orange end points:
- For phenolphthalein: moles of HCl = volume of HCl (in L) × molarity of HCl
- For methyl orange: moles of HCl = volume of HCl (in L) × molarity of HCl
2. Convert the moles of HCl to moles of Na2CO3 and NaHCO3 using the balanced chemical equations:
- Na2CO3 + 2HCl → 2NaCl + H2O + CO2
- NaHCO3 + HCl → NaCl + H2O + CO2
From the balanced equations, we can see that 2 moles of HCl react with 1 mole of Na2CO3 and 1 mole of HCl reacts with 1 mole of NaHCO3.
3. Determine the moles of Na2CO3 and NaHCO3 reacted by comparing the moles of HCl used at each end point with the stoichiometry of the reactions.
4. Calculate the mass of Na2CO3 and NaHCO3 using the moles and molar masses of each compound:
- Mass = moles × molar mass
5. Calculate the percent of Na2CO3 and NaHCO3 in the mixture:
- Percent Na2CO3 = (mass of Na2CO3 / mass of mixture) × 100%
- Percent NaHCO3 = (mass of NaHCO3 / mass of mixture) × 100%
Let's apply these steps to find the percent of Na2CO3 and NaHCO3 in the given mixture.
Given:
- Mass of the mixture (Na2CO3 + NaHCO3) = 0.523 g
- Volume of HCl required for the phenolphthalein end point = 17 mL
- Volume of HCl required for the methyl orange end point = 43.8 mL
- Molarity of HCl = 0.1 M
Step 1:
- Moles of HCl used for phenolphthalein end point = (17 mL / 1000 mL) × 0.1 M = 0.0017 moles HCl
- Moles of HCl used for methyl orange end point = (43.8 mL / 1000 mL) × 0.1 M = 0.00438 moles HCl
Step 2:
- Na2CO3 moles reacted at phenolphthalein end point = 0.0017 moles HCl ÷ 2 = 0.00085 moles Na2CO3
- Na2CO3 moles reacted at methyl orange end point = (0.00438 moles HCl - 0.0017 moles HCl) ÷ 2 = 0.00134 moles Na2CO3
- NaHCO3 moles reacted = 0.0017 moles HCl - 0.00134 moles Na2CO3 = 0.00036 moles NaHCO3
Step 3:
- Mass of Na2CO3 = 0.00085 moles Na2CO3 × molar mass of Na2CO3
- Mass of NaHCO3 = 0.00036 moles NaHCO3 × molar mass of NaHCO3
Step 4:
- Molar mass of Na2CO3 = 46 g/mol
- Molar mass of NaHCO3 = 84 g/mol
Step 5:
- Percent Na2CO3 = (mass of Na2CO3 / mass of mixture) × 100%
- Percent NaHCO3 = (mass of NaHCO3 / mass of mixture) × 100%
By substituting the values from the previous steps, you can calculate the percent of each component in the mixture.