What volume of a 3.5 m hcl is required to completely neutralize 50.0ml of a 2.0 m naoh

Hey there!:

Molarity HCl = 3.5 M

Volume HCl = ?

Molarity NaOH = 2.0 M

Volume NaOH in liters = 50.0 mL / 1000 => 0.05 L

Number of moles NaOH:

n = M * V

n = 2.0 * 0.05

n = 0.1 moles of NaOH

Given the reaction:

HCl + NaOH = NaCl + H2O

1 mole HCl 1 mole NaOH

1 mol HCl reacts with 1 mol NaOH , so moles NaOH = moles HCl

0.1 moles of NaOH = 0.1 moles of HCl

Therefore:

M( HCl ) = n / V

3.5 = 0.1 / V

V = 0.1 / 3.5

V = 0.029 L in mL :  0.029 * 1000 => 29.0 mL

hope that helps!

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