3)A 100 mL sample of hard water is titrated with 22.4 mL of EDTA solution from p

Question

3)A 100 mL sample of hard water is titrated with 22.4 mL of EDTA solution from problem 2. the same amount of MgCl2 is added as previously and the total volume of EDTA solution required is 22.44 mL.

A)What volume of EDTA is used titration the Ca2+ in the hard water?

B)how many moles of EDTA are there in the volume?

C)how many moles of Ca2+ are there in 100 mL of water?

D)assume all of the Ca2+ I'm the water come for CaCo3. how many moles of CaCo3 are in 1L of water?how many grams of CaCo3 are in 1L of water?

E)if 1 ppm CaCo3=1mg/liter, what is the water hardness in ppm CaCO3?

Note:
here are the answers to the first 2 questions if they are needed:

Question 0,00512niol 0,00512mol 250. 1000mL. 0.0205M c) Moles of Ca in 25.0 mL solution 0,00512mol 25.0mL 000512 moi 8.55 Ch Mola of EDIA 25.95m

Explanation / Answer

Solution:

A)

Volume V1 of the sample hard water=100mL

Molarity M1 of the hard water=?

Volume V2 of   EDTA solution =22.44mL

Molarity M2 of EDTA =? =0.0197M (from previous titrations)

We know that

                     V1xM1= V2X M2

Substituting the values we have

                       100mL x? M = 22.44ml x 0.0197M =0.004M

Since we know that the titration reaction of magnesium ion and calcium ion is similar the reactions are as follows

Mg2+ + EDTA4- MgEDTA2-

Ca2+ + EDTA4- CaEDTA2-

Since molarity(moles/L) is same for all the solutions, if we take the same molar solution of Mg2+ and Ca2+ solutions separately for titrations against the EDTA solution of molarity 0/0197M, then the volume of Ca2+ solution required will be the same volume as of Mg2+ solution and the volume of EDTA will be the same 22.44mL

The volume of EDTA is used titration the Ca2+ in the hard water = 22.44mL

B) Molarity of EDTA = 0.0197M

That is 1L (or 1000mL)of EDTA solution has= 0.0197moles

Therefore 22.44mL of EDTA solution has=(0.0197 x 22.44)/1000=0.0004moles

0.0004moles of EDTA are there in the volume

C) Molarity of Ca2+ solution= 0.004M

That is 1L (or 1000mL) of Ca2+ solution has= 0.004moles

Therefore 100mL of Ca2+ solution has= (0.004 x 100)/1000=0.0004moles

0.0004 moles of Ca2+ are there in 100 mL of water

D)      CaCO3                                           Ca2+                  +            CO32-

1mole                                    1mole                            1mole

This means 1mole of Ca2+ comes from 1mole of CaCO3   

Therefore 0.0004moles Ca2+ comes from 0.0004 mole of CaCO3

Molecular mass of     CaCO3 = (40+12+(16x3)=100U

1mole of CaCO3dissolved in 1L of water has = 100g of CaCO3

0.0004 mole of CaCO3dissolved in 1L of water has = (100g x 0.0004)/1of CaCO3

                                                                                   =0.04g     

                 0.0004 moles of CaCO3 are there in 1L of water.

                  0.04 grams of CaCO3 are in 1L of water

   E) 1gram= 1000mg

                   Hence 0.04 grams=1000 x0.04=40mg

                    1 ppm CaCO3=1mg/liter

                40 grams of CaCO3 are in 1L of water =40ppm CaCO3

The hardness of water in ppm CaCO3 =40

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