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50 " mL of " water required 4 " mL of " (N)/(50)HCl for complete neutralisation. 200 " mL of " this water was then boiled with 10 " mL of " (N)/(10) soda reagent. After filtration, the filtrate and the washing were made up to 200 mL with distilled water 50 " mL of " this solution required 8.0 " mL of " (N)/(50) HCl for complete neutralisation. CAlculate the temporary and permanent hardness in ppm.

Answer» <html><body><p></p>Solution :` 4 " mL of " (N)/(50) Hcl=(4xx0.02xx50)/(1000)g CaCO_(3)` <br/> So, the temporary <a href="https://interviewquestions.tuteehub.com/tag/hardness-14971" style="font-weight:bold;" target="_blank" title="Click to know more about HARDNESS">HARDNESS</a>`=(4xx0.02xx50)/(1000xx50)g CaCO_(3)` <br/> `=80ppm` <br/> After boiling with `Na_(2)CO_(3)` and filtration, 8 " mL of " `(N)/(50)` HCl <br/> `-=(8)/(50)=0.16m" Eq of "(HCl)/(50mL) Na_(2)CO_(3)` <a href="https://interviewquestions.tuteehub.com/tag/filtrate-461136" style="font-weight:bold;" target="_blank" title="Click to know more about FILTRATE">FILTRATE</a> <br/> `-=(0.16xx200)/(50)m" Eq of "(HCl)/(200mL) of Na_(2)CO_(3)` filtrate <br/> `-=0.64 mEq` <br/> Initially, total m" Eq of "`Na_(2)CO_(3)-=10xx(1)/(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)=1mEq` <br/> Total mEw of `Na_(2)CO_(3)` reacted `=1-0.64=0.36 mEq` of weight of `Na_(2)CO_(3)=0.36xx10^(-3)xx50g CaCO_(3)` <br/> So <a href="https://interviewquestions.tuteehub.com/tag/permanent-590817" style="font-weight:bold;" target="_blank" title="Click to know more about PERMANENT">PERMANENT</a> hardness`=(0.36xx10^(-3)xx50xx10^(6))/(200)=90ppm` <br/> Hence, temporary hardness`=80ppm`, permanent hardness <br/> `=90ppm`</body></html>


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