What weight of na2co3 of 95% purity would be required to neutralize 45.6ml of 0.235n acid?

I want you step by step clearly explain

answer 0.5978gm

To calculate the weight of Na2CO3 required to neutralize the given quantity of acid, we can follow these steps:

Step 1: Determine the molarity of the acid solution
The given acid has a concentration of 0.235N, which means it contains 0.235 moles of acid in 1 liter of solution.

Step 2: Convert the volume of the acid solution to liters
The given volume of acid is 45.6 mL. We need to convert it to liters by dividing it by 1000:
45.6 mL = 45.6/1000 = 0.0456 L

Step 3: Calculate the number of moles of acid in the given volume
To find the number of moles, we multiply the volume (in liters) by the molarity of the acid:
Number of moles of acid = Volume (in L) x Molarity
Number of moles of acid = 0.0456 L x 0.235 mol/L = 0.010716 mol

Step 4: Determine the molar mass of Na2CO3
The molar mass of Na2CO3 can be found by adding up the atomic masses of its constituent elements: sodium (Na), carbon (C), and oxygen (O).
Na2CO3 = (2 x atomic mass of Na) + (1 x atomic mass of C) + (3 x atomic mass of O)
Na2CO3 = (2 x 22.99 g/mol) + (1 x 12.01 g/mol) + (3 x 16.00 g/mol)
Na2CO3 = 45.98 g/mol + 12.01 g/mol + 48.00 g/mol
Na2CO3 = 105.99 g/mol

Step 5: Determine the molecular weight of the desired Na2CO3 quantity
The molecular weight is the product of the number of moles and the molar mass of Na2CO3:
Weight of Na2CO3 (in g) = Number of moles of acid x Molar mass of Na2CO3
Weight of Na2CO3 = 0.010716 mol x 105.99 g/mol = 1.136 g

Step 6: Adjust for purity
The given Na2CO3 is 95% pure. To find the weight of the Na2CO3 needed, we need to adjust for this:
Weight of Na2CO3 required = (Weight of Na2CO3 calculated) / (Purity)
Weight of Na2CO3 required = 1.136 g / 0.95 = 1.1947 g

Therefore, approximately 1.1947 grams (rounded to four decimal places) of Na2CO3 with 95% purity would be required to neutralize 45.6 mL of 0.235N acid.

To calculate the weight of Na2CO3 required to neutralize the given volume and concentration of acid, you can follow these steps:

Step 1: Determine the molarity of the acid.

Given:
Volume of acid (V) = 45.6 mL
Molarity of acid (M) = 0.235 N

To convert the molarity from normality to moles per liter (M), we need to multiply the normality by the equivalent factor. In this case, the equivalent factor for HCl is 1 because it is monoprotic.

Molarity of acid (M) = Normality of acid (N) × Equivalent factor

M = 0.235 N × 1
M = 0.235 M

Step 2: Convert the volume of the acid from milliliters to liters.

Volume of acid (V) = 45.6 mL = 45.6/1000 = 0.0456 L

Step 3: Use the balanced chemical equation to determine the stoichiometric ratio between the acid and Na2CO3.

2 HCl + Na2CO3 → H2O + CO2 + 2 NaCl

The balanced equation shows that 2 moles of HCl react with 1 mole of Na2CO3.

Step 4: Calculate the number of moles of HCl.

Number of moles (n) = Molarity × Volume
n(HCl) = 0.235 mol/L × 0.0456 L

Step 5: Calculate the number of moles of Na2CO3 using the stoichiometric ratio.

According to the balanced equation, 2 moles of HCl react with 1 mole of Na2CO3.

n(Na2CO3) = (n(HCl) / 2) = (0.235 mol/L × 0.0456 L) / 2

Step 6: Convert the moles of Na2CO3 to grams using the formula weight (atomic weight of Na2CO3).

The formula weight of Na2CO3 = 2(23) + 12 + 3(16) = 106 g/mol

Weight of Na2CO3 = n(Na2CO3) × formula weight of Na2CO3
= [(0.235 mol/L × 0.0456 L) / 2] × 106 g/mol

After calculating, the weight of Na2CO3 required to neutralize the given amount of acid would be approximately 0.5978 grams (rounded to four decimal places).

Therefore, the weight of Na2CO3 of 95% purity required to neutralize 45.6 mL of 0.235 N acid is approximately 0.5978 grams.

mL acid x N acid = milliequivalents acid = 45.6 mL x 0.235 = ? m.e. acid

1 m.e. of anything = 1 m.e. of anything else; therefore, m.e. acid = m.e. Na2CO3.
?m.e. acid = ? m.e. Na2CO3.
m.e. Na2CO3 x milliequivalent weight Na2CO3 = grams Na2CO3
?m.e. Na2CO3 x 0.053 = grams Na2CO3.
Put all of this together to make
mL x N x mew = grams
45.6mL x 0.235N x 0.053g/m.e. =0.56795 g of 100% Na2CO3. Since it is only 95%, then
0.56795/0.95 = 0.5978g which rounds to 0.598g to three significant figures.