If 30 ml of 02 M HCL with 45 ml what is the molarity of NaOH

I think you posted only part of the problem. I assume you have titrated the NaOH with HCl and you want to know the M of the NaOH.

mols HCl = M x L = ?
mols NaOH = mols HCl from the molar ratio of 1:1.
M NaoH = mols NaOH/L NaOH. You know mols NaOH and L NaOH, solve for M NaOH.

I assumed that 45 mL of NaOH, and that "02 M" is 0.02 M HCl.

Write the balanced equation:
HCl + NaOH ---> NaCl + H2O

Then get the moles of HCl by multiply the volume by the molarity:
M = n/V
n = V*M
n = 30 mL * 0.02 M
n = 0.6 mmol HCl

Since in the reaction it's 1 mol HCl reacted for every 1 mol NaOH present, then the number of moles of NaOH is
0.6 mmol HCl * (1 mmol NaOH / 1 mmol HCl) = 0.6 mmol NaOH

Finally, solve for the molarity of NaOH.
M = n/V
M = 0.6 mmol / 45 mL
M = ?

Units in mol/L. Hope this helps~ `u`

To determine the molarity of NaOH, we need to use the concept of stoichiometry and the volume and concentration of HCl.

1. First, calculate the number of moles of HCl using the given volume and concentration.
Moles of HCl = Volume of HCl (in liters) * Concentration of HCl

As the volume of HCl is given in milliliters, we need to convert it to liters:
Volume of HCl = 30 ml = 30/1000 L = 0.03 L

The concentration of HCl is given as 0.2 M (2 M is most likely a typo in the question).

Moles of HCl = 0.03 L * 0.2 M = 0.006 moles of HCl

2. Secondly, we use the balanced chemical equation between NaOH and HCl to determine the molar ratio between NaOH and HCl.

The balanced equation is:
2 NaOH + HCl -> NaCl + H2O

From the equation, we see that 2 moles of NaOH react with 1 mole of HCl.

3. Now, we can determine the number of moles of NaOH using the molar ratio calculated in step 2.

Moles of NaOH = 0.006 moles of HCl * (2 moles of NaOH / 1 mole of HCl) = 0.012 moles of NaOH

4. Finally, we calculate the molarity of NaOH using the volume of NaOH.

Volume of NaOH = 45 ml = 45/1000 L = 0.045 L

Molarity of NaOH = Moles of NaOH / Volume of NaOH
= 0.012 moles / 0.045 L
= 0.267 M

Therefore, the molarity of NaOH is 0.267 M.