100 , mL of a water sample contains 0.81 , g | vedclass

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(A) The hardness of water is expressed in terms of $CaCO_3$ equivalents. $n_{eq}(CaCO_3) = n_{eq}(Ca(HCO_3)_2) + n_{eq}(Mg(HCO_3)_2)$ Given that the n

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$10,000$

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(A) The hardness of water is expressed in terms of $CaCO_3$ equivalents. $n_{eq}(CaCO_3) = n_{eq}(Ca(HCO_3)_2) + n_{eq}(Mg(HCO_3)_2)$ Given that the n-factor for bicarbonate salts is $2$: $\frac{w}{100} 	imes 2 = \frac{0.81}{162} 	imes 2 + \frac{0.73}{146} 	imes 2$ $\frac{w}{50} = 0.005 	imes 2 + 0.005 	imes 2$ $\frac{w}{50} = 0.01 + 0.01 = 0.02$ $w = 0.02 	imes 50 = 1.0 \, g$ Hardness in $ppm = \frac{	ext{Mass of } CaCO_3 	ext{ in } g}{	ext{Mass of water in } g} 	imes 10^6$ Since $100 \, mL$ of water has a mass of $100 \, g$: $	ext{Hardness} = \frac{1.0}{100} 	imes 10^6 = 10,000 \, ppm$

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