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(NONE) The reaction between $Ca(HCO_3)_2$ and $H_2SO_4$ is: $Ca(HCO_3)_2 + H_2SO_4 \rightarrow CaSO_4 + 2H_2O + 2CO_2$.At the equivalence point,the mi

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$400$

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(NONE) The reaction between $Ca(HCO_3)_2$ and $H_2SO_4$ is: $Ca(HCO_3)_2 + H_2SO_4 ightarrow CaSO_4 + 2H_2O + 2CO_2$.At the equivalence point,the milliequivalents of $Ca(HCO_3)_2$ equal the milliequivalents of $H_2SO_4$.$n_{eq}(Ca(HCO_3)_2) = n_{eq}(H_2SO_4) = M 	imes n_{factor} 	imes V(L) = \frac{1}{20} 	imes 2 	imes 0.040 = 0.004 \ eq$.Since $1 \ mol$ of $Ca(HCO_3)_2$ is equivalent to $1 \ mol$ of $CaCO_3$ $(100 \ g/mol)$,the mass of $CaCO_3$ is $0.004 \ mol 	imes 100 \ g/mol = 0.4 \ g$.Hardness in $ppm = \frac{	ext{mass of } CaCO_3 	ext{ in } g}{	ext{volume of water in } mL} 	imes 10^6 = \frac{0.4}{500} 	imes 10^6 = 800 \ ppm$.

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