Given that: NaClO( aq ) + HCl( aq ) → NaCl( aq ) + HClO( aq ), how many mL of 6.00 M HCl is required to neutralize 25.00 mL of 2.35 M NaClO

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Quizplus · Accepted Answer

The reaction between NaClO and HCl is a 1:1 stoichiometry, meaning 1 mole of NaClO reacts with 1 mole of HCl. To find the volume of HCl needed, first calculate the moles of NaClO: $25.00 \, \text{mL} \times \frac{2.35 \, \text{mol}}{1000 \, \text{mL}} = 0.05875 \, \text{mol}$. Since the reaction is 1:1, $0.05875 \, \text{mol}$ of HCl is also needed. Using the concentration of HCl (6.00 M), the volume of HCl required is $ \frac{0.05875 \, \text{mol}}{6.00 \, \text{M}} = 0.0097917 \, \text{L}$, or $9.79 \, \text{mL}$.

Given That: NaClO( Aq ) + HCl( Aq ) →