Question

1. what volume of 0.496 m hcl is required to neutralize 20.0 ml of 0.809 m sodium hydroxide?

Explanation:

Step1: Write neutralization reaction

HCl + NaOH → NaCl + H₂O (1:1 mole ratio)

Step2: Calculate moles of NaOH

Moles = Molarity × Volume = \(0.809\ \text{mol/L} × 0.0200\ \text{L} = 0.01618\ \text{mol}\)

Step3: Find moles of HCl needed

Moles of HCl = Moles of NaOH = \(0.01618\ \text{mol}\)

Step4: Calculate volume of HCl

Volume = \(\frac{\text{Moles}}{\text{Molarity}} = \frac{0.01618\ \text{mol}}{0.496\ \text{mol/L}} ≈ 0.0326\ \text{L} = 32.6\ \text{mL}\)

Answer:

32.6 mL