1.A corrosion technologist pipetted a 100.00 ml hard water sample and titrated i

Question

1.A corrosion technologist pipetted a 100.00 ml hard water sample and titrated it with 37.64 ml of 0.01 M EDTA solution for a total hardness endpoint, and 29.32 ml of 0.01 M EDTA solution for a calcium endpoint. Calculate the concentration of magnesium and calcium in ppm. Record your answers as ppm Mg2+, and Ca2+ respectively. For each answer indicate the ppm as CaCO3 also.

2.Write and balance the reaction between the magnesium ion, the bicarbonate ion and heat.

3.Write and balance the scale removal reaction involving the scale generated in question #2 with HCl.

4.In a sour water system (H2S) there is an opportunity for ferrous sulphide to form along with the calcium and magnesium carbonate scales. If acidizing is used to clean away scales, what danger exists? Indicate this by using a balanced reaction.

5.In the field, where the scale in question #4 is removed by acidizing, what precautions should be taken?

Explanation / Answer

Ca and Mg ions complexed by EDTA to form CaEDTA 2- and MgEDTA 2-

1)

a) Technologist required 37.64 ml of 0.01 M EDTA for 100 ml hard water to reach total hardness endpoint.

In total hardness endpoint, both ions Ca and Mg are complexed.

The balanced chemical equation for these reactions is

Ca2+ + H2EDTA = CaEDTA2- + 2H+

Mg2+ + H2EDTA = MgEDTA2- + 2H+

Let us consider M2+ as all metal ions in solution.

M2+ + H2EDTA2- = MEDTA2- + 2H+

As we can see, one mole of Metal ion(M2+) reacts with 1 mole of EDTA2-

Therefore, moles M2+ = moles of EDTA2-

Now we will calculate moles of EDTA = concentration * Volume(L)

                                                                   = 0.01 * 37.64 * 10^-3 (ml to L)

                                                                         = 0.0003764 moles

As, moles M2+ = moles of EDTA2-

Moles of M2+ = 0.0003764 moles which comprises both (Ca2+ and Mg2+)

b)

For, calcium endpoint, 29.32 ml 0.01 M EDTA was used. This considers Calcium only.

Now we will calculate moles of EDTA = concentration * Volume(L)

                                                                   = 0.01 * 29.32 * 10^-3 (ml to L)

                                                                         = 0.0002932 moles

As, moles Ca2+ = moles of EDTA2-

Moles of Ca2+ = 0.0002932 moles in 100 ml = 0.002932 moles/ L

Both solutions were same, therefore we can subtract moles of Ca2+ from total endpoint value to get Mg2+ moles.

Moles of Mg2+ = 0.0003764 – 0.0002932 = 0.0000832 moles in 100 ml = 0.000832 moles/L

c) To convert moles/L to ppm. We have to multiply by its atomic weight.

Ca2+ = atomic weight of Ca2+ ion is 40.08

0.002932 * 40.08 = 0.1175 g/L

Now, we convert gram to mg = 117.5 mg/L = 117.5 ppm

Mg2+ = 0.0000832 * 24.305 = 0.0020 g/L

Now, we convert gram to mg = 2 mg/L = 2 ppm

d)

In case of ppm as CaCO3 we consider all Metal ions as Calcium to get total CaCO3.

Therefore, ppm as CaCO3 would be = 117.5 + 2 = 119.5 ppm

2)

Mg^2+ + 2HCO3- =heat= MgCO3 + CO2 + H2O

3)

MgCO3 + 2HCl = MgCl2 + CO2 + H2O

4) Ferrous sulphide is pyrophoric which can ignite in air spontaneously.

And with reaction of HCl it produces H2S, which is very toxic gas.

FeS + 2HCl = FeCl2 + H2S

5) As ferrous sulphide can react with air we have to carry out scale removal in anaerobic condition if possible.

Also, proper PPE(Protective Personal Equipment) such as gas mask should be used while scale removal as H2S is toxic.

People working in that environment can get accustomed to H2S smell, therefore outside person or H2S sensors should be used.

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