Question

1 concentration of EDTA solution 2 molar mass of CaCO3 0.0100 M 100.09 g/mol Trail 1 -Trail 2 Trail 3 12 12.01 11.99 5 Unknown analysis Part 1 (pH 10)_ 6 volume of unknown, mL 7 initial volume of EDTA in buret 8 final volume of EDTA in buret 9 net volume of EDTA used 0 1.01 1.02 1 Data analysis for part 1:1 mol EDTA = 1 mol metal ions moles of EDTA used (MXV, L) moles of (Mg2+ + Ca2+) concentration of (Mg2+ + Ca2+) in sample, M concentration of (Mg2+ + Ca2+) in sample, mmol/L total water hardness in ppm = mg/L CaCO3 16 17 18 19 part 2 - titration at pH 12.5 with ca Trail 1 Trail 2 Trail 3 20 volume of unknown, mL 12 12.01 11.99 21 initial volume of EDTA in buret 010 22 final volume of EDTA in buret 0.89 0.91 0.9 23 net volume of EDTA used 24 Data analysis for part 2: 1 mole EDTA = 1 mol metal ions moles of EDTA used ( MXV, L) moles of (Ca2+) concentration of Ca2+) in sample, M concentration of (Ca2+ ) in sample, mmol/L concentration of (Mg2+) in sample, mmol/L

how would I calculate my results from the first two data the first data is my results and the second data I have to calculate each of those with the information that I have on the first 3 trails I really really need help and i just dont understand thank you.

Answer 1

PART 1

Average volume of unknown sample taken = (12.0 + 12.01 + 11.99) / 3 = 12.0 mL = 12 x 10-3 L

Average Net volume of EDTA used = (1.01 + 1.02 + 1.0)/ 3 = 1.01 mL = 1.01 x 10-3 L

moles of EDTA used = volume of EDTA used x molarity of EDTA solution = 1.01 x 10-3 x 0.0100 = 1.01 x 10-5 moles.

Moles of (Mg2+ + Ca2+) = moles of EDTA used = 1.01 x 10-5 moles.

Concentration of (Mg2+ + Ca2+) in sample, M = 1.01 x 10-5 moles/ 12 x 10-3 L = 0.0842 x 10-2 moles/ L

= 8.42 x 10-4 M.

Concentration of (Mg2+ + Ca2+) in sample, mmol/L = 0.0842 x 10-2 moles/10-3/ L = 0.842 mmol/ L.

Now molar mass of CaCO3 = 100.09 g/mol = 100.09 mg/ mmol

Therefore, Total hardness of water in ppm = mg/ L of CaCO3 = 0.842 x 100.09 mg/ L of CaCO3

= 84.28 mg/ L of CaCO3.

PART 2

Average volume of unknown sample taken = (12.0 + 12.01 + 11.99) / 3 = 12.0 mL = 12 x 10-3 L

Average Net volume of EDTA used = (0.89+ 0.91 + 0.90)/ 3 = 0.90 mL = 0.90 x 10-3 L

moles of EDTA used = volume of EDTA used x molarity of EDTA solution = 0.90 x 10-3 x 0.0100 = 9 x 10-6 moles.

Moles of Ca2+= moles of EDTA used = 9 x 10-6 moles.

Concentration of Ca2+ in sample, M = 9 x 10-6 moles/ 12 x 10-3 L = 0.75 x 10-3 moles/ L = 7.5 x 10-4 M.

Concentration of Ca2+ in sample, mmol/L = 0.75 x 10-3 moles/10-3/ L = 0.75 mmol/ L.

Concentration of Mg2+ in sample, mmol/L =

Concentration of (Mg2+ + Ca2+) in sample(from Part 1) - Concentration of Ca2+ in sample

= 0.842 -0.75 = 0.092 mmol/ L.