how many ml of 0.15 m naoh solution are required to neutralize 35.00 ml of 0.22 m hcl

To determine the number of milliliters (mL) of 0.15 M NaOH solution required to neutralize 35.00 mL of 0.22 M HCl solution, we need to use the concept of stoichiometry and the balanced chemical equation of the reaction between NaOH and HCl.

The balanced chemical equation for the reaction is as follows:
NaOH + HCl -> NaCl + H2O

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

Now, let's calculate the moles of HCl in 35.00 mL of 0.22 M HCl solution:
moles of HCl = volume (in L) × molarity
= 0.035 L × 0.22 mol/L
= 0.0077 mol

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

Now, let's calculate the volume of 0.15 M NaOH solution required to neutralize 0.0077 moles of HCl:
volume (in L) = moles ÷ molarity
= 0.0077 mol ÷ 0.15 mol/L
= 0.0513 L

To convert the volume into milliliters:
volume (in mL) = volume (in L) × 1000
= 0.0513 L × 1000
= 51.3 mL

Therefore, 51.3 mL of 0.15 M NaOH solution is required to neutralize 35.00 mL of 0.22 M HCl solution.