What volume of a 0.125 M solution of NaOH is needed to neutralize 25.0 mL of a 0.285 M solution of HCL

To find the volume of the NaOH solution needed to neutralize the HCl solution, we can use the concept of stoichiometry and the balanced chemical equation between NaOH and HCl.

The balanced chemical equation between NaOH and HCl is:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

From this equation, we can see that the stoichiometric ratio of NaOH to HCl is 1:1. This means that 1 mole of NaOH reacts with 1 mole of HCl.

First, we need to determine the number of moles of HCl in the given 25.0 mL of the 0.285 M HCl solution.

Molarity (M) is defined as moles of solute per liter of solution.

We can use the formula:
Molarity = moles of solute / volume of solution (in liters)

Rearranging the formula, we have:
moles of solute = Molarity × volume of solution (in liters)

Converting the volume from mL to liters:
25.0 mL = 25.0 / 1000 = 0.025 L

Plugging in the values into the formula:
moles of HCl = 0.285 M × 0.025 L = 0.007125 moles

Since the stoichiometric ratio is 1:1, we know that it will take the same number of moles of NaOH to neutralize the HCl.

Now, we need to find the volume of the 0.125 M NaOH solution required to react with 0.007125 moles of NaOH.

Using the same formula:
moles of solute = Molarity × volume of solution (in liters)

Rearranging the formula to solve for the volume of the solution:
volume of solution (in liters) = moles of solute / Molarity

Plugging in the numbers:
volume of solution (in liters) = 0.007125 moles / 0.125 M = 0.057 L

Since the volume is in liters, we can convert it back to milliliters by multiplying by 1000:
volume of solution (in mL) = 0.057 L × 1000 = 57.0 mL

Therefore, 57.0 mL of the 0.125 M NaOH solution is needed to neutralize 25.0 mL of the 0.285 M HCl solution.

v * .125 = 25.0 mL * .285