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Determine the volume of 0.150 M NaOH solution required to neutralize 145 mL of a 0.25 M HCl solution. The neutralization reaction is NaOH(aq) + HCl(aq) → H 2 O(l) + NaCl(aq)

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Answer:-

241.66 mL

Explanation:-

Volume of HCl = 145 mL / 1000 = 0.145 L

Strength of HCl = 0.25 M

Number of moles of HCl = Strength x volume

= 0.25 M x 0.145 L

= 0.03625 moles

According to the balanced chemical equation

NaOH(aq) + HCl(aq) → H 2 O(l) + NaCl(aq)

1 mol of HCl reacts with 1 mol of NaOH.

∴0.03625 moles of HCl reacts with 0.03625 moles of NaOH.

Strength of NaOH = 0.15 M

Volume of NaOH = Number of moles / strength

= 0.03625 moles / 0.15 M

= 0.24166 L x 1000

= 241.66 mL

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