University:
Massachusetts Institute of Technology
Course:
1.76 | Aquatic Chemistry
Academic year:

2021
Views:

262

Pages:

17
Author:

esfredint7qo

PROBLEM SET #4 1.76: Aquatic PROBLEM 1 (a) KAlSi₃O₈ (s) → Al(OH)₃ (s) + 3H₂SiO₄ *balance Si with 3H₂SiO₄ *balance O with 4H₂O *balance H with H⁺ *balance K with K⁺ KAlSi₃O₈ (s) + 4H₂O + H⁺ = Al(OH)₃ (s) + 3H₂SiO₄ + K⁺ (b) Book rxn: ½ Albite + 3H⁺ + H₂O = ½ Na-mont + H₂SiO₃ + 3 Na⁺ Re-written so Na⁺, H⁺ have coefficients of 1: Multiply above rxn by 2/3 ⅔ Albite + ⅔ H⁺ + ⅔ H₂O = ⅓ Na-mont + ⅔ H₂SiO₃ + 2 Na⁺ Which reduces to ⅔ Albite + H⁺ + 5/9 H₂O = ⅓ Na-mont + 5/9 H₂SiO₃ + 2 Na⁺ 9) Using information in Table S.4 ½ (Albite + H⁺ + 5/9 H₂O = ½ Kaolinite + 2 H₂SiO₄ + Na⁺) log K = 0.168 Y₁ (Na-Mont + H⁺ + H₂O = 1/2 Dol + 1/2 H₂SiO₃ + 3 Na⁺) log K = -9.1 Multiply Y₁ by -4 and add to rxn 1: ½ Albite + H⁺ + 5/9 H₂O - 4/3 Na-Mont - 4/3 H₂SiO₃ - 4 Na⁺ = ½ Kaolinite + 2 H₂SiO₄ + Na⁺ - 2/3 Dol - 2/3 H₂SiO₃ - 4 Na⁺ ½ Albite + H⁺ + 5/9 H₂O = ½ Kaolinite + 4/3 H₂SiO₄ + Na⁺ - 2/3 Dol as desired. K_eq = 10^0.723 = [H₂SiO₄]^4/3 [Na⁺] / [H⁺] d) Assume [Na⁺] = [Na⁺] = 10⁻² M Given: [H₂SiO₄] = [CH₄SiO₄] + [H₃SiO₄⁻] + [SiO₃²⁻] K₁: [H₃SiO₄⁻] = 10⁻⁹.86 [H₂SiO₄] / [H⁺] K₂: [SiO₃²⁻] = 10⁻¹³.¹ [H₃SiO₄⁻] / [H⁺] = 10⁻¹³.¹ (10⁻⁹.⁸⁶ [H₂SiO₄] / [H⁺]) / [H⁺] = 10⁻²³¹ [H₂SiO₄] / [H⁺]² [HCO₃⁻] = (CH₄SiO₄)K₁K₂ [H⁺]² [CO₃²⁻] = (CH₄SiO₄)K₁K₂K₃ [H⁺]³ [OH⁻] = Kw / [H⁺] = 10⁻¹⁴ / [H⁺] pH = 8.48 Cₜ = 10⁻².⁷² M Alk = 10⁻².⁷² - 2g₁ C₁ = 10⁻³.¹⁷ M Alk Tableau H₂O H⁺ CaCO₃ H₂CO₃ H₂SiO₃ Na⁺ H₂O OH⁻ HCO₃⁻ CO₃²⁻ Ca²⁺ H₂SiO₃ HSiO₃⁻ SiO₃²⁻ Na⁺ Alk = -T_OTH = -[H⁺] + [OH⁻] + [HCO₃⁻] - 2[CO₃²⁻] - 2[Ca²⁺] + [H₂SiO₃]

1.76 Problem Set 4 Solutions

Related Documents

Recommended Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock