If 400 mL of 3.5 molar HCl solution is added to 200 mL of 2.3 molar NaOH solution, what will be the molarity of NaCl in the resulting solution? Answer in units of M.

This is a limiting reagent.

HCl + NaOH ==> NaaCl + H2O

mols HCl = M x L = ? = approx 1.4
mols NaOH = M x L = ? = approx 0.46
So the amount of NaCl formed is 0.46 mols. The volume is 400 + 200 = 600 mL = 0.6 L
M = mols/L

To determine the molarity of NaCl in the resulting solution, first we need to find the moles of HCl and NaOH.

Step 1: Calculate the moles of HCl.
Moles of HCl = Volume of HCl solution (in L) x Molarity of HCl
= 400 mL (converted to L) x 3.5 M
= 0.4 L x 3.5 M
= 1.4 moles

Step 2: Calculate the moles of NaOH.
Moles of NaOH = Volume of NaOH solution (in L) x Molarity of NaOH
= 200 mL (converted to L) x 2.3 M
= 0.2 L x 2.3 M
= 0.46 moles

Step 3: Determine the limiting reactant.
To determine the limiting reactant, we compare the moles of HCl and NaOH. The reactant that is completely consumed is the limiting reactant.

Based on the stoichiometry of the reaction, 1 mole of HCl reacts with 1 mole of NaOH to form 1 mole of NaCl.

Since we have 1.4 moles of HCl and 0.46 moles of NaOH, NaOH is the limiting reactant because it is present in a lower amount.

Step 4: Calculate the moles of NaCl formed.
Moles of NaCl = Moles of NaOH (since NaOH is the limiting reactant)

Moles of NaCl = 0.46 moles

Step 5: Calculate the volume of the resulting solution.
The total volume of the resulting solution is the sum of the volumes of the HCl and NaOH solutions.

Total volume = Volume of HCl solution + Volume of NaOH solution
= 400 mL (converted to L) + 200 mL (converted to L)
= 0.4 L + 0.2 L
= 0.6 L

Step 6: Calculate the molarity of NaCl.
Molarity (M) = Moles of solute / Volume of solution (in L)

Molarity of NaCl = Moles of NaCl / Volume of resulting solution
= 0.46 moles / 0.6 L
≈ 0.767 M

Therefore, the molarity of NaCl in the resulting solution is approximately 0.767 M.

To find the molarity of NaCl in the resulting solution, we need to use the concept of stoichiometry, which is the quantitative relationship between reactants and products in a chemical reaction.

In this case, the reaction between HCl and NaOH can be represented as follows:
HCl + NaOH -> NaCl + H2O

It is important to note that the number of moles of HCl is equal to the number of moles of NaCl produced in this reaction. Therefore, we can use the molarity and volume to determine the number of moles of HCl and NaCl.

First, we calculate the number of moles of HCl:
Moles of HCl = Molarity of HCl × Volume of HCl
Moles of HCl = 3.5 M × 0.4 L (since 400 mL is equal to 0.4 L)
Moles of HCl = 1.4 moles

Next, we calculate the number of moles of NaCl produced:
Since the reaction between HCl and NaOH is 1:1, the number of moles of NaCl will also be 1.4 moles.

Now, we calculate the total volume of the resulting solution:
Total volume = Volume of HCl + Volume of NaOH
Total volume = 0.4 L + 0.2 L (since 200 mL is equal to 0.2 L)
Total volume = 0.6 L

Finally, we calculate the molarity of NaCl in the resulting solution:
Molarity of NaCl = Moles of NaCl ÷ Total volume
Molarity of NaCl = 1.4 moles ÷ 0.6 L
Molarity of NaCl ≈ 2.33 M

Therefore, the molarity of NaCl in the resulting solution is approximately 2.33 M.