Ass 3_sol

# Ass 3_sol - CIVL 242 Assignment 3 Solution Question 1 Cl2 H...

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1 CIVL 242 Assignment 3 Solution Question 1: +- 22 Cl H O HOCl+H Cl +⇔ + formula [ ] [] 4 2 H Cl HOCl 41 0 Cl K ⎡⎤ ⎣⎦ == × HOCl H OCl ⇔+ formula 7.58 HO C l 10 HOCl a K According to mass balance, [ ] - 5000 / Cl HOCl OCl Cl 0.00704 71000 / total remain mg L M mg mol =+ + = = Since more than 99% of Cl 2 hydrolyzes, [ ] 2 Cl remain is so low that it can be neglected. Therefore, [ ] [ ] - 2 HOCl OCl Cl 0.00704 / total mol L +≈ = eq. + log H pH =− +6 H1 0 1 0 pH M −− - 7.58 1.58 6 + OCl 10 10 HOCl 10 H a K M M = From eq. and eq. , [ ] HOCl and - OCl can be solved. [ ] - HOCl 0.0686 , OCl 0.0018 M M Actually, there is a little Cl 2 remaining in the solution, which does not react with water. [ ] [ ] [ ] [ ] - 2 2 Cl Cl Cl Cl Cl 0.0704 react total remain total M ≈= [ ] 6 5 2 4 HC l H O C l 10 0.0704 0.0686 Cl 1.21 10 0 remain M K ×× = × ×

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2 Therefore, after dissolution and hydrolysis of 5000mg/L of Cl 2 . [ ] 2 Cl is 5 1.21 10 M × , [ ] HOCl is 0.0686 M , - OCl ⎡⎤ ⎣⎦ is 0.0018 M Question 2: 2+ - 2 Ca(OCl) Ca 2OCl ⇔+ [] 4 2 40 / Ca(OCl) 2.797 10 143000 / mg L M mg mol == × +- HOCl H OCl 7.58 HO C l 10 HOCl a KM
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Ass 3_sol - CIVL 242 Assignment 3 Solution Question 1 Cl2 H...

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