High School

What volume (in mL) of a 0.244 M NaOH solution is required to titrate 50.0 mL of 0.210 M H\(_2\)SO\(_4\)?

A. 116 mL
B. 86.1 mL
C. 97.4 mL
D. 58.1 mL
E. 43.0 mL

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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