Answer: Given 30 mL of 0.05 M EDTA and 25 mL is Ca+2. The balanced equation betw

ID: 700203 • Letter: A

Question

Answer:

Given 30 mL of 0.05 M EDTA and 25 mL is Ca+2.

The balanced equation between Ca+2 and EDTA is

Ca^+2 + EDTA -------->[Ca(EDTA)]^-2.

1 mole of Ca+2 react with 1 mol of EDTA

Moles of EDTA=Molarity xvolume=0.05 mol/Lx0.030 L

=0.0015 mol EDTA.

Moles of Ca+2=0.0015 mol

Concentration of Ca^+2=moles/volume=0.0015 mol/0.025 L= 0.06 M.

It means 0.06 moles of Ca^+2 in 1 L water.

Molar mass of Ca+2=40 g/mol.

Mass of Ca^+2=molesxmolar mass=0.06 molx40 g/mol

Mass of Ca^+2=2.4 g=2400 mg. (1g=1000 mg)

We know that 1ppm=1 mg in 1 L solution.

Here we have 2400 mg of Ca^+2 in 1 L water.

So concentration of Ca^+2=2400 ppm.

2+ water. Calculate the concentration of Ca2 in ppm. (The atomic weight of Calcium is 40.078) (3 marks)

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