What is the amount of 4M HCl needed to completely neutralize 50mL of 2M NaOH

mols NaOH = M x L = 2 x 0.05 = 0.10

mols HCl must be 0.10
Then M HCl = mols HCl/L HCl.
L HCl = mols HCl/M HCl = 0.10/4M = ?

According to the equation HCl+NaOH = NaCl+H2O it will take just as many moles of HCl as NaOH to do the job.

So, now just do the math. If it will take x mL, then

4x = 2*50

To determine the amount of 4M HCl needed to completely neutralize 50mL of 2M NaOH, you can use the concept of stoichiometry and the equation of the reaction between HCl and NaOH.

The balanced chemical equation for the reaction between HCl and NaOH is:

HCl + NaOH -> NaCl + H2O

According to the stoichiometry of the reaction, one mole of HCl reacts with one mole of NaOH to produce one mole of NaCl and one mole of water.

First, we need to determine the number of moles of NaOH present in 50mL of a 2M solution. The number of moles can be calculated using the formula:

moles = concentration (in M) x volume (in L)

Converting 50mL to L:

volume (in L) = 50mL / 1000 = 0.05 L

Now, substituting the given values into the formula:

moles of NaOH = 2M x 0.05 L = 0.1 moles

Since the stoichiometry of the reaction is 1:1 between HCl and NaOH, we would need an equal number of moles of HCl to neutralize the NaOH.

Therefore, the amount of 4M HCl needed can be calculated using the formula:

volume (in L) = moles / concentration (in M)

Substituting the values:

volume (in L) = 0.1 moles / 4M = 0.025 L

To express this answer in milliliters (mL), we can convert it by multiplying by 1000:

volume (in mL) = 0.025 L x 1000 = 25 mL

So, you would need 25mL of 4M HCl to completely neutralize 50mL of 2M NaOH.