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Answer 1 · 2017-05-02T18:48:41

A hardness of 320.7 ppm corresponds to 0.3556 g of $"CaCl"_2$ per litre.

A hardness of 1 ppm refers to the concentration of $"Ca"^"2+"$ expressed as 1 ppm of $"CaCO"_3$ .

$"1 ppm of hardness" = "1 g"/(10^6color(white)(l) "g") = "1 mg"/(10^3color(white)(l) "g") = "1 mg/L"$

∴ $"320.7 ppm of hardness" = "320.7 mg/L" = "0.3207 g/L"$

Thus, you need 0.3207 g of $"CaCO"_3"$ per litre.

For $"CaCl"_2, M_text(r) = 110.98$ .

For $"CaCO"_3, M_text(r) = 100.09$ .

Thus, to get the same amount of $"Ca"^"2+"$ , you need

$0.3207 color(red)(cancel(color(black)("g CaCO"_3))) × "110.98 g CaCl"_2/(100.09 color(red)(cancel(color(black)("g CaCO"_3)))) = "0.3556 g CaCl"_2$