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Grade 12Physical Chemistry

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.

Profile image of Shirish Chandra Srivastava
10 Years agoGrade 12
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1 Answer

Profile image of Vikas TU
9 Years ago
Write the total milliequivalents of CO32- in 100 ml as given:
that would be: 100/50 = 2.
After neutralizing 25 ml,
meq. of co3 gets remained = (200/25)*8.2/50 = 1.3
thus CACO3 resent in 100 ml water is = (2-1.3)*50 = 35 gm.
permanent hardness therefore becomes = 35/200 * 1000 = 175 ppm.