Question

What volume (mL) of 0.125 M NaOH is required to neutralize 12.4 mL of 0.369 M HCl? What volume (mL) of 0.125 M NaOH is required to neutralize 12.4 mL of 0.369 M HCl? 36.6 4.20 0.572 0.000572 0.0273

Answer 1

Answer:

There is needed 36.6 mL of 0.125M NaOH to neutralize the 12.4 mL 0.369 M HCl

Explanation:

Step 1: The balanced equation

NaOH + HCl ⇔ NaCl + H2O

Sodium hydroxide (NaOH) and hydrochloric acid (HCl) react in a  1:1 mole ratio to produce aqueous sodium chloride and water.

This means 1 mole of NaOH is needed when 1 mole of HCl is consumed to produce 1 mole of NaCl and 1 mole of water.

Step 2: Calculate The required volume NaOH to neutralize  HCl

Because the mole ratio is 1:1 we will use the following formula

C1*V1 = C2*V2

with C1 = the concentration of NaOH = 0.125 M

with V1 = the volume of NaOH = TO BE DETERMINED

with C2 = the concentration of HCl = 0.369M

with V2 = the volume of HCl = 12.4 mL = 0.0124 L

0.125M * V1 = 0.369M * 0.0124 L

V1 = (0.369 * 0.0124)/ 0.125M = 0.0366 L = 36.6 mL

There is needed 36.6 mL of 0.125M NaOH to neutralize the 12.4 mL 0.369 M HCl