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Volume of 0.150 M NaOH solution required to neutralize 25mL of a 0.055 M HCl solution?
Equation of the reaction: NaOH(aq) + HCl(aq) ——-> H2O(l) + NaCl(aq)

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nkirotedoreen46

Answer:

9.17 mL

Explanation:

In this question we are given;

  • Molarity of NaOH as 0.150 M
  • Molarity of the acid, HCl as 0.055 M
  • Volume of the acid, HCl as 25 mL
  • Equation for the reaction as;

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

We are required to determine the volume of the alkali, NaOH used;

Step 1: Determine the moles of the acid, HCl

Molarity = Number of moles ÷ Volume

Thus, rearranging the formula;

Moles = Molarity × volume

          = 0.055 M × 0.025 L

          = 0.001375 moles

Step 2: Moles of the alkali, NaOH

  • From the reaction; 1 mole of HCl reacts with 1 mole of NaOH

Therefore;

  • Moles of HCl = Moles of NaOH

Hence; Moles of NaOH = 0.001375 moles

Step 3: Determine the volume of the alkali, NaOH

  • From the previous equation;

Molarity = Moles ÷ Volume

  • Rearranging the equation;

Volume = Moles ÷ Molarity

Therefore;

Volume of NaOH = 0.001375 moles ÷ 0.150 M

                             = 0.00917 L, but, 1 L = 1000 mL

                             = 9.17 mL

Therefore, the volume of NaOH required is 9.17 mL

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