Question

A 100.00 mL sample of spring water was treated to convert any iron present to Fe^2+. Addition of 25.00 mL of 0.002107 M K_2 Cr_2 O_7 resulted in the reaction 6Fe^2+ + Cr_2O_7^2- + 14 H_+ rightarrow 6Fe^3+ + 2Cr^3+ + 7H_2O The excess K_2 Cr_2 O_7 was titrated with 7.47 mL of a 0.00979 M Fe^2+ solution. Calculate the ppm of iron in the sample.

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Answer 1

We have to convert the 25.00mL of 0.002107M K₂Cr₂O₇ to moles (remember to use volume in liters):

total moles = M*L = (0.002107 mol/L)(0.025L)

total moles = 0.000052675 moles K₂Cr₂O₇

Those are the total moles of K₂Cr₂O₇ used in the titration, however there was an excess which was titrated with Fe²⁺, we have to calcultate the moles of K₂Cr₂O₇ in excess so we can know the moles of K₂Cr₂O₇ which reacted with iron in the water sample:

We calculate the Fe²⁺ moles that reacted with the excess K₂Cr₂O₇ :

moles Fe²⁺ = M*L

moles Fe²⁺ = (0.00979 mol/L) (0.00747L)

moles Fe²⁺ = 0.0000731313 moles Fe²⁺

Now, using the coefficients in the balanced chemical equation, we calculate the excess moles of K₂Cr₂O₇

excess moles K₂Cr₂O₇ = 0.0000731313 moles Fe²⁺ (1 mol K₂Cr₂O₇ / 6 mol Fe²⁺)

excess moles K₂Cr₂O₇ = 0.00001218855 moles K₂Cr₂O₇

Now we calculate the moles of K₂Cr₂O₇ that reacted with iron in the water, it's just a substraction:

total moles K₂Cr₂O₇ - excess moles K₂Cr₂O₇ = reaction moles K₂Cr₂O₇

reaction moles K₂Cr₂O₇ = 0.00004048645 moles K₂Cr₂O₇

We convert those moles to moles of Fe²⁺ present in the sample, again using the coefficients in the balanced equation:

moles Fe²⁺ (sample) = 0.00004048645 moles K₂Cr₂O₇ (6 mol Fe²⁺ / 1mol  K₂Cr₂O₇)

moles Fe²⁺ (sample) = 0.0002429187 moles Fe²⁺

To calculate ppm we convert those moles to miligrams (using molar mass of Fe = 55.845g/mol) and multiplying by 1000 (to convert to miligrams) and divide by sample volume (0.1L)

miligrams Fe²⁺ = 0.0002429187 moles Fe²⁺ (55.845g / mol) (1000mg/1g)

miligrams Fe²⁺ = 13.566 mg

ppm = mg/L

ppm = 13.6mg/0.1L

ppm = 136ppm Fe²⁺