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InterviewSolution
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A 200 mL sample of hard water requires 33.0 " mL of " 0.01 M `H_2SO_4` for complete neutralisation. 200 " mL of " the same sample was boiled with 15.0 " mL of " 0.1 M NaOH solution, filtered and made up to 200 mL again. This sample now requires 53.6 " mL of " 0.01 M `H_2SO_4`. Calculate Mg hardness. |
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Answer» 33.0 " mL of " `0.01xx2NH_2SO_4` `-=(333.0xx0.01xx2xx50)/(1000)` `-=0.033gCaCO_3(EwCaCO_3=50)` `-=(0.033xx10^(6))/(200mL)-=165ppm of CaCO_3` Eq Thus temporary hardness `-=165ppm` `mEqNaOH=15xx0.1xx1=1.5` `m" Eq of "H_2SO_4=53.6xx0.01xx2=1.072` `m" Eq of "NaOH` (excess) that reacted with water `=1.5-1.072=0.428mEq` `0.428mEq NaOH-=(0.428xx50xx10^(6))/(1000xx200)ppmCaCO_3` `-=107ppm of CaCO_3` Thus Mg hardness `-=107ppm` |
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