What is the normality and nature of a mixture obtained by mixing 0.62 g of Na2CO3 •H2O to 100mL of 0.1N H2SO4?
To determine the normality and nature of the mixture obtained by mixing 0.62 g of Na2CO3 •H2O to 100 mL of 0.1N H2SO4, we can follow these steps:
Step 1: Calculate the number of moles of Na2CO3 •H2O.
To calculate the number of moles, we need to divide the mass of Na2CO3 •H2O by its molar mass.
The molar mass of Na2CO3 •H2O = (2 * atomic mass of Na) + atomic mass of C + (3 * atomic mass of O) + (2 * atomic mass of H) + atomic mass of O
= (2 * 23) + 12 + (3 * 16) + (2 * 1) + 16
= 46 + 12 + 48 + 2 + 16
= 124 g/mol
Number of moles of Na2CO3 •H2O = mass / molar mass
= 0.62 g / 124 g/mol
= 0.005 moles
Step 2: Calculate the volume of 0.1N H2SO4 required for the reaction.
To calculate the volume of 0.1N H2SO4, we can use the equation M1V1 = M2V2, where M1 is the concentration of the Na2CO3 •H2O and V1 is the volume, and similarly for H2SO4.
Molarity of Na2CO3 •H2O = Number of moles / Volume in liters
= 0.005 moles / 0.1 liters
= 0.05 N
Since the reaction between Na2CO3 •H2O and H2SO4 occurs in a 1:1 ratio, the normality of the mixture will be 0.05 N.
Step 3: Determine the nature of the mixture.
Na2CO3 •H2O is a basic compound, and H2SO4 is an acidic compound. When these two compounds react, a neutralization reaction takes place, resulting in the formation of H2O and a salt. In this case, the salt formed is Na2SO4.
Therefore, the nature of the mixture obtained by mixing 0.62 g of Na2CO3 •H2O to 100 mL of 0.1N H2SO4 is neutral, and its normality is 0.05 N.
To determine the normality and nature of the mixture obtained by mixing Na2CO3 •H2O and H2SO4, we need to understand the concept of normality and perform some calculations.
Normality is a measure of concentration that accounts for the chemical reaction that the substance undergoes when it reacts with another substance. It is expressed in terms of equivalents of solute per liter of solution.
First, we need to calculate the number of equivalents of Na2CO3 •H2O and H2SO4 separately.
1. Calculate the equivalents of Na2CO3 •H2O:
- We have 0.62 g of Na2CO3 •H2O.
- Calculate the molar mass of Na2CO3 •H2O:
Na = 22.99 g/mol, C = 12.01 g/mol, O = 16.00 g/mol, H = 1.01 g/mol
Na2CO3 •H2O = (2 * 22.99) + 12.01 + (3 * 16.00) + (2 * 1.01) + 16.00
= 105.99 g/mol
- Calculate the number of moles of Na2CO3 •H2O:
Moles = Mass / Molar mass = 0.62 g / 105.99 g/mol
≈ 0.00584 mol
- Calculate the equivalents:
Equivalents = Moles * Number of equivalents
= 0.00584 mol * 2
≈ 0.0117 N
2. Calculate the equivalents of H2SO4:
- We have 100 mL (0.1 L) of 0.1 N H2SO4.
- The normality of the given H2SO4 is already expressed, so no conversion is needed.
- Therefore, the equivalents of H2SO4 is 0.1 N.
Next, we need to determine the nature of the mixture:
- If the equivalents of the acid (H2SO4) are equal to or greater than the equivalents of the base (Na2CO3 •H2O), the nature of the mixture is acidic.
- If the equivalents of the base (Na2CO3 •H2O) are greater, it will be either basic or neutral.
Comparing the equivalents obtained:
- Equivalents of Na2CO3 •H2O: 0.0117 N
- Equivalents of H2SO4: 0.1 N
Since the equivalents of H2SO4 are greater than Na2CO3 •H2O, the nature of the mixture will be acidic.
Therefore, the normality of the mixture obtained by mixing 0.62 g of Na2CO3 •H2O with 100 mL of 0.1 N H2SO4 is 0.1 N and the nature of the mixture is acidic.
I don't understand the problem.
Na2CO3 + H2SO4 ==> Na2SO4 + H2O + CO2
There is the equation. I have no idea what "nature of the mixture" means but let's see what happens.
How many grams Na2CO3 will the H2SO4 use? That's ml x N x milliequivalent weight Na2CO3 = grams Na2CO3.
100 mL x 0.1N x 0.053 = approx 0.52 but you need a more accurate answer than that.
You had 0.62 g Na2CO3 initially, you have used 0.52 which leaves about 0.1 g Na2CO3 unused. (Remember to recalculate all of these numbers.) The normality of that solution will be N = eq/L
equivalents = grams/eq weight = 0.1/53 = ?
N = eq/L = ?eq/0.1L = ?
So the nature of the solution may be that the H2SO4 is completely used and you have a solution of Na2CO3 in the liquid and the N of the Na2CO3 solution is ?? N.