Answer :
Final answer:
To titrate 50.0 mL of 0.210 M H2SO4 with a 0.244 M NaOH solution, approximately 86.1 mL of the NaOH solution is required.
Explanation:
To determine the volume of a 0.244 M NaOH solution required to titrate 50.0 mL of 0.210 M H2SO4, we can use the balanced chemical equation for the reaction between NaOH and H2SO4:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
From the equation, we can see that the stoichiometric ratio between H2SO4 and NaOH is 1:2. This means that for every 1 mole of H2SO4, 2 moles of NaOH are required.
Given that the concentration of the H2SO4 solution is 0.210 M and the volume used is 50.0 mL (0.050 L), we can calculate the number of moles of H2SO4:
moles of H2SO4 = concentration × volume = 0.210 M × 0.050 L = 0.0105 mol
Since the stoichiometric ratio is 1:2, we need twice as many moles of NaOH:
moles of NaOH = 2 × 0.0105 mol = 0.021 mol
Now, we can use the concentration of the NaOH solution (0.244 M) to calculate the volume:
volume of NaOH = moles of NaOH / concentration = 0.021 mol / 0.244 M ≈ 0.0861 L = 86.1 mL
Therefore, the volume of 0.244 M NaOH solution required to titrate 50.0 mL of 0.210 M H2SO4 is approximately 86.1 mL.
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