Question

17.94 mL of HCl is required to fully react with 1.401 g of sodium carbonate (Na2CO3, MW = 106g/mol). What is the molarity of the HCl?

Answer 1

The molarity of the HCl solution is 0.735 M.

Step-by-step explanation:

To determine the molarity of the HCl solution, we need to use the given information about the volume of HCl solution and the number of moles of sodium carbonate.

Given:

Volume of HCl solution = 17.94 mL

Mass of sodium carbonate (Na2CO3) = 1.401 g

Molar mass of sodium carbonate (Na2CO3) = 106 g/mol

We can use the formula:

Molarity (M) = moles of solute / volume of solution (in L)

First, we need to calculate the number of moles of sodium carbonate:

Moles of Na2CO3 = mass / molar mass

= 1.401 g / 106 g/mol

= 0.0132 mol

Since the reaction between HCl and sodium carbonate has a 1:1 stoichiometry, the number of moles of HCl required will also be 0.0132 mol.

Now, we can calculate the molarity of the HCl solution:

Molarity = moles of HCl / volume of HCl solution (in L) = 0.0132 mol / 0.01794 L

= 0.735 M