An aqueous calcium carbonate standard solution is known to have a hardness of 12

Question

An aqueous calcium carbonate standard solution is known to have a hardness of 128.0 ppm (hardness due to CaCO3). Suppose 50.00 mL of the calcium carbonate standard solution is passed through a filter and the filtrate collected (assume no solution volume was lost in the filter). The filtered solution is found to have a hardness of 32.0 ppm due to CaCO3. Assume that the density of the solutions is 0.9980 g/mL and the molar mass of calcium carbonate is 100.0 g/mol. Based on this information, answer the following questions (a-c). Be sure to show your calculations.

a. What is the efficiency of the filter for removing hardness due to calcium ions from the standard solution? Express your answer as a percentage.

b. How many moles of calcium ions are present in 50.00 mL of the filtered solution?

c. If the 50.00 mL of filtered solution is titrated with 0.00250 M EDTA solution, what volume (in milliliters) of the EDTA solution is needed to reach the equivalence point?

Explanation / Answer

An aqueous calcium carbonate standard solution is known to have a hardness of 128.0 ppm (hardness due to CaCO3). Suppose 50.00 mL of the calcium carbonate standard solution is passed through a filter and the filtrate collected (assume no solution volume was lost in the filter). The filtered solution is found to have a hardness of 32.0 ppm due to CaCO3. Assume that the density of the solutions is 0.9980 g/mL and the molar mass of calcium carbonate is 100.0 g/mol. Based on this information, answer the following questions (a-c). Be sure to show your calculations.

a. What is the efficiency of the filter for removing hardness due to calcium ions from the standard solution? Express your answer as a percentage.

Hardness before filtration = 128.0 ppm

Hardness after filtration = 32.00 pm

efficiency of the filter for removing hardness = 128-32/128*100

= 96/128*100

= 72.7 %

b. How many moles of calcium ions are present in 50.00 mL of the filtered solution?

The hardness of filtrate solution = 32.0 ppm.

ppm = mg/L

32 ppm = 32.0 mg/ L

Volume = 50.00 ml or 0.050 L

so, how many mg would be there in 0.05L

32.00 mg x 0.05L/1L = 1.6mg

Now, convert mg into gram

1.6 mg x 1g/1000mg = 0.0016 g

Then divide these grams by molar mass:

0.0016 g/ 100.0869 g/mol

= 1.6*10^-5 moles CaCO3

c. If the 50.00 mL of filtered solution is titrated with 0.00250 M EDTA solution, what volume (in milliliters) of the EDTA solution is needed to reach the equivalence point?

Number of moles of CaCO3 = 1.6*10^-5 moles CaCO3

H2Y2- + Ca2+ CaY2- + 2H+

n = 1.6*10^-5
C = 0.00250
V = ? in L

V = n / C = 1.6*10^-5 / 0.00250 moles / L

V= 6.4*10^-3 L

V = 6.4 mL

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