Question

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.

Answer 1

                  1.998×10^-3mol   X mol

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

                                         2 mol        1 mol

C₁(OH^-) = K_B – 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^-) = K_B – 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)×V_M = 9.99×10^-4 mol × 22.4 L/mol = 2.2378×10^-2 L = 22.378 mL ≈ 22.38 mL.