Answer to Question #95200 in General Chemistry for Mike

Question #95200
2Na+2H2O ---> 2NaOH +H2
While investigating this reaction velocity we also measured solution's pH, which change showed in graph. Calculate released hydrogen H2 volume by normal conditions after 500s (pH12) from experiment begining (which was pH9). The resulting aqueous NaOH solution's volume was 200ml. Give the consistent answer.
1
Expert's answer
2019-09-25T05:43:47-0400

                  1.998×10-3mol   X mol

2Na + 2H2O --->                2NaOH   +    H2

                                         2 mol        1 mol


C1(OH-) = KB – С1(H+) = 10-14M2 / 10-9M = 10-5 M.

n1(OH-) = C1(OH-) × V = 10-5 M × 0.2 L = 2 × 10-6 mol.

C2(OH-) = KB – С2(H+) = 10-14M2 / 10-12M = 10-2 M.

n2(OH-) = C2(OH-) × V = 10-2 M × 0.2 L = 2 × 10-3 mol.

∆n(NaOH) = ∆n(OH-) = n2(OH-) – n1(OH-) = 2 × 10-3 mol – 2 × 10-6 mol = 1.998 × 10-3 mol.

n(H2) = X = 1.998×10-3 / 2 mol = 9.99×10-4 mol.

V(H2) = n(H2)×VM = 9.99×10-4 mol × 22.4 L/mol = 2.2378×10-2 L = 22.378 mL ≈ 22.38 mL.



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