20 mL of a solution of HCl concentration, neutralized 0,50 g of Na2CO3. What is the Molarity of the acid solution?
To find the molarity of the acid solution, we need to determine the number of moles of HCl that neutralized the given amount of Na2CO3.
The first step is to calculate the number of moles of Na2CO3:
1. Use the molecular weight of Na2CO3 to convert the mass to moles:
Molecular weight of Na2CO3 = 22.99 g/mol (Na) + 2 * 16.00 g/mol (O) + 12.01 g/mol (C)
= 22.99 g/mol + 32.00 g/mol + 12.01 g/mol
= 85.00 g/mol
Moles of Na2CO3 = Mass of Na2CO3 / Molecular weight of Na2CO3
= 0.50 g / 85.00 g/mol
≈ 0.00588 mol
The second step is to determine the number of moles of HCl:
The balanced chemical equation for the reaction between HCl and Na2CO3 is:
2HCl + Na2CO3 -> 2NaCl + H2O + CO2
From the equation, we can see that 2 moles of HCl react with 1 mole of Na2CO3. So, in this case, the number of moles of HCl will be twice the number of moles of Na2CO3, as the reaction is stoichiometrically balanced.
Therefore, Moles of HCl = 2 * Moles of Na2CO3
= 2 * 0.00588 mol
= 0.01176 mol
Finally, we can calculate the molarity of the acid solution:
Molarity (M) = Moles of solute / Volume of solution (in liters)
= Moles of HCl / Volume of solution (in liters)
The volume of the solution is given as 20 mL, which is equivalent to 0.020 L.
Molarity (M) = 0.01176 mol / 0.020 L
= 0.588 M
Therefore, the molarity of the acid solution is 0.588 M.