A 50.00-mL sample of groundwater is titrated with 0.0900 M EDTA. Assume that Ca2 accounts for all of the hardness in the groundwater. If 11.40 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 CaCO3 by mass?
2 answers
5415
Moles of EDTA = 0.001026moles
The end point is where the moles of EDTA and moles of Ca2+ have to be the same. Hence, moles of Ca2+ is also 0.001026moles.
There's a 1:1 ration of Ca2+ to CaCO3, so CaCO3, also have 0.001026moles.
The molarity of CaCO3 is (0.001026mols)/(0.05000L)= 0.02052M CaCo3.
Molarity = moles/L, PPM = mg/L
so (0.02052mole CaCO3/1L)(100.1g/1 mole CaCO3)(1000mg/1g)((1ppm x 1L)/1mg) = 2052ppm
The end point is where the moles of EDTA and moles of Ca2+ have to be the same. Hence, moles of Ca2+ is also 0.001026moles.
There's a 1:1 ration of Ca2+ to CaCO3, so CaCO3, also have 0.001026moles.
The molarity of CaCO3 is (0.001026mols)/(0.05000L)= 0.02052M CaCo3.
Molarity = moles/L, PPM = mg/L
so (0.02052mole CaCO3/1L)(100.1g/1 mole CaCO3)(1000mg/1g)((1ppm x 1L)/1mg) = 2052ppm