ter is due to the p is titrated with EDTA, a chelating agent, in the presence of

Question

ter is due to the p is titrated with EDTA, a chelating agent, in the presence of the indicator eriochrome black T as In. E blue when Ca? is removed. ized here black T, a weaker chelating agent than EDTA, is red in the presence of Ca2 and tuns red A 50.00-mL sample of groundwater is titrated with 0.0150 M EDTA. the hardness in the groundwater. If 11.60 mL of EDTA is required to titrate the 50.00-mL sample, what is the hardness of the groundwater in molarity and in parts per million of CaCOs by mass? M CaCO ppm CaCo

Explanation / Answer

11.60 mL of 0.015 M EDTA titrates 50 mL of groundwater sample.

11.60 mL of 0.015 M EDTA is equal to 11.60 mg of CaCO3

Amount of CaCO3 in 50 mL groundwater = 11.60 mg

Amount of CaCO3 in 1000 mL groundwater = 232 mg/L

In ppm:

Amount of CaCO3 in 100 mL groundwater = 23.2 mg

Amount of CaCO3 in 100 g groundwater = 232 mg

Amount of CaCO3 in 106 g groundwater = 23.2 x 10-3 x 106/100 = 232 ppm

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