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)

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