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
