Hi please help with this.

You will mix 5 ml of each 2M HCL and 2M NaOH solutions, calculate the concentration of NaCl formed in the new solution?

moles HCl = M x L = 2 x 0.005.

moles NaOH = M x L = 2 x 0.005.
moles NaCl = moles HCl = moles NaOH
M NaCl = mols/L = 0.01mole/0.010 L = 1.0M

thank you for the help, one more question how did you get the volume for NaCl?

You had 5 mL of HCl and you added 5 mL of NaOH which makes 10 mL total volume and that is 0.010 L. The NaCl that was formed was formed IN this volume.

To calculate the concentration of NaCl formed in the new solution, we need to determine the number of moles of NaCl present by using the stoichiometry of the balanced chemical equation.

The balanced chemical equation for the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is:

HCl + NaOH → NaCl + H2O

From the equation, we can see that one mole of HCl reacts with one mole of NaOH to form one mole of NaCl.

First, let's calculate the number of moles of HCl and NaOH used. We have 5 ml of each 2M HCl and 2M NaOH solutions, which means:

Number of moles of HCl = concentration (M) × volume (L)
= 2M × 0.005L
= 0.01 moles of HCl

Number of moles of NaOH = concentration (M) × volume (L)
= 2M × 0.005L
= 0.01 moles of NaOH

Since the stoichiometry of the reaction is 1:1, the number of moles of NaCl formed will be the same as the number of moles of HCl and NaOH used.

Therefore, the concentration of NaCl in the new solution is:

Concentration of NaCl = moles of NaCl / volume of solution

Since the volumes of HCl and NaOH are mixed, the total volume of the solution will be 5 mL + 5 mL = 10 mL = 0.01 L.

Concentration of NaCl = 0.01 moles / 0.01 L
= 1 M

So, the concentration of NaCl formed in the new solution is 1 M.