what weight of na2c03 of 95% purity would be required to neutralize 45.6ml of 0.235N acid?
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mL x N x mew = grams (at 100%)
Then grams/0.95 = grams Na2CO3 at 95%.
Note: mew Na2CO3 = molar mass/2
To find the weight of Na2CO3 required to neutralize the given volume and concentration of acid, you need to follow these steps:
Step 1: Determine the molar concentration of the acid solution.
The concentration of the acid is given as 0.235N (normality). However, since Na2CO3 is a diprotic base, we need to convert this to molarity (M). The conversion factor for a diprotic acid is 2, so the molar concentration of the acid is 0.235N / 2 = 0.1175 M.
Step 2: Write the balanced chemical equation for the neutralization reaction.
The balanced equation for the reaction between sodium carbonate (Na2CO3) and the acid is:
2 HX + Na2CO3 -> 2 NaX + H2O + CO2
Where HX represents the acid.
Step 3: Calculate the number of moles of HX in the given volume of acid solution.
To calculate the number of moles (n) of HX, use the formula:
n = M * V
Where M is the molar concentration of the acid and V is the volume of the acid in liters.
In this case, V is given as 45.6 mL, which is equivalent to 0.0456 L.
So, n = 0.1175 M * 0.0456 L = 0.00535 moles.
Step 4: Determine the number of moles of Na2CO3 required.
Referring to the balanced equation, we see that for every 2 moles of HX, we need 1 mole of Na2CO3. Therefore, the number of moles of Na2CO3 required is half the number of moles of HX.
0.00535 moles HX * 0.5 = 0.002675 moles Na2CO3.
Step 5: Calculate the molecular weight of Na2CO3.
The molecular weight of Na2CO3 is calculated by adding the atomic masses of its components: 2 * (Na) + (C) + 3 * (O).
2 * (22.99 g/mol) + 12.01 g/mol + 3 * (16.00 g/mol) = 105.99 g/mol.
Step 6: Calculate the weight of Na2CO3 needed.
To determine the weight of Na2CO3 needed, multiply the number of moles of Na2CO3 by its molecular weight:
0.002675 moles Na2CO3 * 105.99 g/mol = 0.2838 grams.
Therefore, approximately 0.2838 grams of Na2CO3 of 95% purity would be required to neutralize 45.6 mL of 0.235N acid.