200 ml of hard water was boiled with 100 ml N/50 Na2CO3 reagent. After boiling, the volume was again made to 200 ml and the solution filtered. 25 ml of the filterate required 8.2 ml of N/50 HCl for neutralization. Calculate the permanent hardness in ppm.

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Tushar · Answer

Milliequivalents of CO3 ions in 100 ml N/50 Na2CO3 solution= 100*(1/50) = 2 As solution was again made 200ml and 25 ml was neutralized was 8.2ml N/50 HCl, mEq of CO3 ions remaining in the filterate = (200/25)*8.2*1/50 = 1.3 mEq Therefore mEq of CO3 ions precipitated as CaCO3 = 2-1.3= 0.7 As equivalent weight of CaCO3 is 50, amount of CaCO3 present in 200ml water = 0.7*50= 35gm Therefore, permanent hardness= (35/200)*1000= 175ppm

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200 ml of hard water was boiled with 100 ml N/50 Na2CO3 reagent. After boiling, the volume was again made to 200 ml and the solution filtered. 25 ml of the filterate required 8.2 ml of N/50 HCl for neutralization. Calculate the permanent hardness in ppm.

200 ml of hard water was boiled with 100 ml N/50 Na2CO3 reagent. After boiling, the volume was again made to 200 ml and the solution filtered. 25 ml of the filterate required 8.2 ml of N/50 HCl for neutralization. Calculate the permanent hardness in ppm.

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200 ml of hard water was boiled with 100 ml N/50 Na2CO3 reagent. After boiling, the volume was again made to 200 ml and the solution filtered. 25 ml of the filterate required 8.2 ml of N/50 HCl for neutralization. Calculate the permanent hardness in ppm.

200 ml of hard water was boiled with 100 ml N/50 Na2CO3 reagent. After boiling, the volume was again made to 200 ml and the solution filtered. 25 ml of the filterate required 8.2 ml of N/50 HCl for neutralization. Calculate the permanent hardness in ppm.

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