0.5g of $CaCo_3$ was dissolved in dilute HCL and diluted to 500 ml with distilled water ,50 ml of this solution required 48 ml of EDTA solution for titration.50ml of hard water sample required 15 ml of EDTA solution for titration.50 ml of same water sample on boiling and filtering requires 10 ml of EDTA solution . Calculate temporary, permanent and total hardness in ppm.

Mumbai University > First Year Engineering > Sem1 > Applied Chemistry 1

Marks: 5M

Year: May 13, Dec 13

Concentration of S.H.W. $= 0.5 \ \ gm \ \ \ CaCO_3/ 500 ml D.W. \\ = 500 \ mgs \ \ in \ \ 500 \ \ ml \ \ water \\ = 1 mg/ml$

50 ml SHW required = 48 ml of EDTA solution

∴ 1 ml EDTA solution = 50/48 mgs $CaCO_3$ equivalent hardness.

50 ml water sample = 15 ml EDTA solution.

∴ hardness of sample = (15 * 50/48) mgs CaCO3 equivalents for 50 ml sample.

Hardness per litre of sample = (15 * 50/48)* 1000/50 mgs/lit.

Total hardness = 312.50 ppm

50 ml water sample after boiling = 10 ml EDTA solution

Permanent hardness of sample = 10 * 50/48 mgs $CaCO_3$ equivalent for 50 ml.

Permanent hardness of one litre sample = (10 * 50/48)*1000/50 mgs/lit.

Permanent hardness of sample = 208.33 ppm

Temporary hardness = total hardness – permanent hardness= 312.5 – 208.33 = 104.17