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Question

# A water contains the following dissolved ions. [Na+]=56mg/L;[Ca2+]=40mg/L; [Mg2+]=30mg/L;[Al3+]=3mg/L; [HCO−3]=190mg/L;[Cl−]=165mg/L; Water pH is 7. Atomic weights: Ca=40,Mg=24,Al=27,H=1,C=12,O=16,Na=23,Cl=35.5 The non-carbonate hardness of the sample in mg/L as CaCO3 is

A
225
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B
156
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C
zero
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D
86
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Solution

## The correct option is D 86Total hardness is due to multivalent metallic cations (Ca2+,Mg2+,Al3+,Fe2+,Mn2+,Sr2+..etc) as CaCO3 Total hardness = conc. of Ca2+ ×Eq.wt.ofCaCO3Eq.wt.ofCa2++conc.ofMg2+ ×Eq.wt.ofCaCO3Eq.wt.ofMg2++conc.ofAl3+× Eq.wt.ofCaCO3Eq.wt.ofAl3+ =40×5020+30×5012+3×50(27/3) =241.66mg/L as CaCO3 Alkalinity is due to CO−3 and HCO−3 of Ca2+ and Mg2+ Alkalinity = conc. of HCO−3×Eq.wt.ofCaCO3Eq.wt.ofHCO−3 =190×5061 =155.73mg/L as CaCO3 Carbonate Hardness = Minimum of total hardness and alkalinity =155.73mg/L as CaCO3 Non carbonate hardness =Total hardness - Carbonate hardness =241.56−155.73 =85.93 =86mg/L as CaCO3

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