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Question

Chlorine gas (8 mg/L as Cl2) was added to a drinking water sample. If the free chlorine residual and pH was measured to be 2 mg/L (as Cl2 ) and 7.5, respectively, what is the concentration of residual OCl− ions in the water? Assume that the chlorine gas added to the water is completely converted to HOCl and OCl−. Atomic weight of Cl = 35.5 Given, OCl−+H+k⇌ HOCl K=107.5

A
5.634×105mol/L
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B
2.817×103mol/L
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C
1.408×105mol/L
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D
1.127×104mol/L
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Solution

The correct option is C 1.408×10−5mol/LpH=7.5 [H+]=10−pH =10−7.5 Free residual chlorine (HOCl+OCl−) =2 mg/L as Cl2 =2×10−3g/L as Cl2 =2×10−3molecular weight of Cl2 mol/L [HOCl]+[OCl−]=2×10−371mol/L ⇒[HOCl]=2×10−371−[OCl−] Given, OCl−+H+k⇌HOCl K=107.5 K=[HOCl][H+][OCl−] ⇒107.5=2×10−371[OCl−]10−7.5[OCl−] 2×[OCl−]=2×10−371 ⇒[OCl−]=1.408×10−5mol/L

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