Answer:
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
