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10 mL of 0.01 (N) HCl is required for ti...

10 mL of 0.01 (N) HCl is required for titrating 100 mL of a sample of cold water using methyl orange as indicator. Determine the temporary hardness of that sample of water.

Text Solution

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10 mL 0.01 (N) HCl solution `equiv` 1 mL 0.1 (N) HCl.
1 g equivalent HCl `equiv` 1 g equivalent `"CaCO"_(3)`
`therefore` 1 mol 0.1 (N) HCl `equiv 50 xx (1)/(100) xx 0.1g " CaCO"_(3)`
`" "` = 0.005 g `"CaCO"_(3)`
Therefore, in 100 mL or 100g of that sample of water contains some hardness-producing substance which is equivalent to 0.005 g `CaCO_(3)`
`therefore " "10^(6)` g of water contains the hardness-producing
subtance equivalent to `(0.005xx10^(6))/(10^(2))` = 50 g `"CaCO"_(3)`.
Hence, the hardness of that sample of water is = 50 ppm.
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