If it takes 54 mL of 0.1 NaOH to neutralize 125 mL of an HCL solution, what is the concentration of HCL?

0.0432 M

Explanation:

We are given;

Volume of NaOH as 54 mL

Molarity of NaOH as 0.1 M

Volume of HCl as 125 mL

We are required to determine the concentration of HCl

Step 1; We write a balanced equation for the reaction between NaOH and HCl

The balanced equation is given by;

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

Step 2: Determine the number of moles of NaOH

Moles = Volume × molarity

Therefore;

Moles of NaOH = 0.054 L × 0.1 M

                          = 0.0054 Moles

Step 3: We use the mole ratio to determine the moles of HCl

From the equation;

1 mole of NaOH reacts with 1 mole of HCl

Therefore;

Moles of NaOH = Moles of HCl

Thus; moles of HCl = 0.0054 moles

Step 4: Determine the concentration of HCl

We know that;

Molarity = Moles ÷ Volume

Therefore;

Molarity of HCl = 0.0054 moles ÷ 0.125 L

                         = 0.0432 M

Therefore, the concentration of HCl is 0.0432 M

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