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?

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

To find the hardness of the groundwater in molarity, we can use the stoichiometry of the reaction between EDTA and Ca2+.

The balanced chemical equation for the reaction is:
Ca2+ + EDTA → Ca-EDTA

From the equation, we can see that 1 mole of Ca2+ reacts with 1 mole of EDTA.

Given that 11.40 mL of 0.0900 M EDTA is required to titrate the sample, we can calculate the number of moles of EDTA used:
moles of EDTA = volume × concentration
moles of EDTA = 0.01140 L × 0.0900 mol/L
moles of EDTA = 0.001026 mol

Since the stoichiometry of the reaction is 1:1 between Ca2+ and EDTA, the number of moles of Ca2+ present in the sample is also 0.001026 mol.

Now, we can calculate the molarity of Ca2+ in the groundwater sample:
molarity of Ca2+ = moles of Ca2+ / volume of sample
molarity of Ca2+ = 0.001026 mol / 0.0500 L
molarity of Ca2+ = 0.0205 M

To find the hardness of the groundwater in parts per million (ppm) of CaCO3 by mass, we can use the conversion factor that 1 ppm = 1 mg/L.

The molar mass of CaCO3 is 100.1 g/mol, which means that 1 mg/L of CaCO3 is equivalent to 0.01 mg/L of Ca.

Therefore, the hardness of the groundwater in ppm of CaCO3 can be calculated as:
hardness (ppm) = molarity of Ca2+ × molar mass of Ca × 1000
hardness (ppm) = 0.0205 mol/L × 40.08 g/mol × 1000
hardness (ppm) = 819.24 ppm

So, the hardness of the groundwater is 0.0205 M in molarity and 819.24 ppm of CaCO3 by mass.

To find the hardness of the groundwater in molarity (M), we can use the equation:

M1V1 = M2V2

Where:
M1 = concentration of the EDTA solution
V1 = volume of the EDTA solution
M2 = concentration of the calcium ion (Ca2+)
V2 = volume of the groundwater sample

Given:
M1 = 0.0900 M (concentration of the EDTA solution)
V1 = 11.40 mL (volume of the EDTA solution)
V2 = 50.00 mL (volume of the groundwater sample)

Rearranging the equation, we have:

M2 = (M1 * V1) / V2

M2 = (0.0900 M * 11.40 mL) / 50.00 mL

M2 = 0.02052 M

Therefore, the hardness of the groundwater is 0.02052 M.

To find the hardness in parts per million (ppm) of CaCO3 by mass, we need to convert the molarity to ppm.

The conversion from molarity to ppm is given by the equation:

ppm = (molarity * molar mass) / 1000

The molar mass of CaCO3 is:
Ca = 40.08 g/mol
C = 12.01 g/mol
O = 16.00 g/mol (3 times)

Molar mass of CaCO3 = (40.08 + 12.01 + 16.00 * 3) g/mol = 100.09 g/mol

ppm = (0.02052 M * 100.09 g/mol) / 1000

ppm = 2.058 g/L

Since 1 L of water has a mass of 1000 g, we can convert g/L to mg/L:

ppm = 2.058 g/L * 1000 mg/g = 2058 mg/L

Therefore, the hardness of the groundwater is 2058 ppm of CaCO3 by mass.

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