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In the titration of 35.0ml of drain cleaner that contains NaOH, 50.08mL of 0.409M HCl must be added to reach the equivalence point. What is the concentration of the base in the cleaner?

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Answer:

The concentration of NaOH in the drain cleaner is approximately 0.586 mol/L.

Explanation:

To find the concentration of the base (NaOH) in the drain cleaner, we can use the concept of stoichiometry and the equation for the balanced chemical reaction between NaOH (base) and HCl (acid):

NaOH + HCl → NaCl + [tex]H_{2}O[/tex]

The balanced equation indicates a 1:1 mole ratio between NaOH and HCl.

  • Calculate the moles of HCl used:

            Moles of HCl = Molarity × Volume

            Moles of HCl = 0.409 mol/L × 0.05008 L ≈ 0.02050mol

  • Use the mole ratio to find moles of NaOH:

            Since the reaction is 1:1, the moles of NaOH will be the same as the moles of HCl.

             Moles of NaOH ≈ 0.02050mol

  • Calculate the concentration of NaOH in the drain cleaner:

            Concentration of NaOH = Moles of NaOH​ / Volume of Drain Cleaner (in liters)

             Volume of Drain Cleaner = 0.0350 L

             Concentration of NaOH = Moles of NaOH / 0.0350 L

             Concentration of NaOH = 0.02050mol / 0.0350 L ≈ 0.5857mol/L

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