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 (L)
Moles of NaOH = $0.809\ \text{M} × 0.0200\ \text{L} = 0.01618\ \text{mol}$

Step3: Relate moles of HCl to NaOH

Moles of HCl = Moles of NaOH = $0.01618\ \text{mol}$

Step4: Calculate volume of HCl

Volume = $\frac{\text{Moles}}{\text{Molarity}}$
Volume of HCl = $\frac{0.01618\ \text{mol}}{0.498\ \text{M}} ≈ 0.0325\ \text{L} = 32.5\ \text{mL}$

Answer:

32.5 mL