Question

Hi, I need help in these two questions, mostly on setting up the equations. Also, how different is it that this question is asking for molarity of "sodium chromate soln" and not "sodium dichromate soln."? thanks in advance!

1. Chromate and ferrous ions react in acidic solution to form chromic and ferric ions,
respectively. If 0.9422 grams of iron(II) bisulfate dissolved in sulfuric acid solution
requires 24.83 mL of sodium chromate solution for complete titration, what is the
molarity of the sodium chromate solution? Show the balanced net ionic and complete
chemical equations for this reaction.

2. The sodium chromate solution from problem 1 was used to titrate a solution made by
dissolving 1.9642 g of a pure ferrous salt in sulfuric acid. The titration required 32.73 mL
of the sodium chromate solution. Calculate the percent by mass of iron in the pure salt.
The net ionic equation here is the same as in problem 1. You will not be able to write a
balanced molecular equation for this because the anion in the ferrous salt was not specified in this problem.

Answer 1

Q1. Sodium chromate is Na₂CrO₄

Sodium dichromate is : Na₂Cr₂O₇

For this question, sodium chromate Na₂CrO₄ is used.

Balanced net ionic equation : 3 Fe²⁺ + CrO₄^2- + 8 H⁺$\rightarrow$ 3 Fe³⁺ + Cr³⁺ + 4 H₂O

mass Fe(HSO₄)₂ = 0.9422 g

moles Fe(HSO₄)₂ = (mass Fe(HSO₄)₂) / (molar mass Fe(HSO₄)₂)

moles Fe(HSO₄)₂ = (0.9422 g) / (250 g/mol)

moles Fe(HSO₄)₂ = 3.769 x 10^-3 mol

initial moles Fe²⁺ = moles Fe(HSO₄)₂

initial moles Fe²⁺ = 3.769 x 10^-3 mol

moles CrO₄^2- used = (initial moles Fe²⁺) * (1 mole CrO₄^2- / 3 moles Fe²⁺)

moles CrO₄^2- used = (3.769 x 10^-3 mol) * (1 / 3)

moles CrO₄^2- used = (3.769 x 10^-3 mol) * (0.333)

moles CrO₄^2- used = 1.256 x 10^-3 mol

volume CrO₄^2- used = 24.83 mL = 0.02483 L

concentration Na₂CrO₄ = (moles CrO₄^2- used) / (volume CrO₄^2- used in Liter)

concentration Na₂CrO₄ = (1.256 x 10^-3 mol) / (0.02483 L)

concentration Na₂CrO₄ = 0.05060 M

Q2.

volume Na₂CrO₄ used = 32.73 mL = 0.03273 L

moles CrO₄^2- used = (concentration Na₂CrO₄) * (volume Na₂CrO₄ used in Liter)

moles CrO₄^2- used = (0.05060 M) * (0.03273 L)

moles CrO₄^2- used = 1.656 x 10^-3 mol

moles Fe²⁺ present in salt = (moles CrO₄^2- used) * (3 moles Fe²⁺ / 1 mole CrO₄^2-)

moles Fe²⁺ present in salt = (1.656 x 10^-3 mol) * (3 / 1)

moles Fe²⁺ present in salt = (1.656 x 10^-3 mol) * (3)

moles Fe²⁺ present in salt = 4.968 x 10^-3 mol

mass Fe²⁺ present in salt = (moles Fe²⁺ present in salt) * (molar mass Fe)

mass Fe²⁺ present in salt = (4.968 x 10^-3 mol) * (55.845 g/mol)

mass Fe²⁺ present in salt = 0.2774 g

percent by mass of iron = (mass Fe²⁺ present in salt / mass of salt) * 100

percent by mass of iron = (0.2774 g / 1.9642 g) * 100

percent by mass of iron = (0.1412) * 100

percent by mass of iron = 14.12 %