Consider the acid/base properties of iron(iii). a. find the ph of a solution made by dissolving 0.7438 g iron(iii) chloride hexahydrate in 200.00 ml of water using [fe(h2o)6] 3+ (aq) ⇌ [fe(h2o)5oh] 2+ (aq) + h+ (aq) pka = 2.83 b. the iron(iii) can be made to precipitate from the solution by addition of hydroxide to form iron(iii) hydroxide. at what ph will the iron(iii) concentration be reduced to 1.0×10−10 m?

a. 2.29

b. 1.56

Explanation:

a. For the reaction given, let's make an equilibrium chart. But first, let's calculate the initial concentration of iron(III) chloride hexahydrate. The molar masses are:

Fe = 55.8 g/mol

H = 1 g/mol

O = 16 g/mol

Fe(H₂O)₆ = 55.8 + 6*2*1 + 6*16 = 163.8 g/mol

n = mass/ molar mass

n = 0.7438/163.8

n = 0.0045 mol

The initial concentration is n/V(L) = 0.0045/0.2 = 0.0225 M

[Fe(H₂O)₆]³⁺(aq) ⇄ [Fe(H₂O)₅OH]²⁺(aq) + H⁺(aq)

0.0225                           0                           0           Initial

-x                                   +x                          +x           Reacts ( stoichiometry 1:1:1)

0.0225 - x                      x                            x            Equilibrium

Ka is the equilibrium constant, and:

Ka =

Ka =

Ka = 1.479x10⁻³

And

1.479*10⁻³ = (x*x)/(0.0225 - x)

1.479*10⁻³ = x²/(0.0225 - x)

x² = 3.32775*10⁻⁵ - 1.479*10⁻³x

x² + 1.479*10⁻³x - 3.32775*10⁻⁵

By Bhaskara's equation:

Δ = (1.479*10⁻³)² - 4*1*(-3.32775*10⁻⁵)

Δ = 1.353x10⁻⁴

x = (-1.479*10⁻³±√1.353x10⁻⁴)/2

x must be positive, so let's ignore the negative value:

x = (-1.479*10⁻³ +√1.353x10⁻⁴)/2

x = 5.076*10⁻³

So, [H⁺] = 5.076*10⁻³

pH = -log[H⁺]

pH = -log(5.076*10⁻³)

pH = 2.29

b. If the iron(III) concentration reduces to 1.0x10⁻¹⁰ M, it means that the reaction with the hydroxide consumed:

0.0225 - 1.0x10⁻¹⁰ = 0.022499 M which will form more iron (II) and H⁺. So the new concentration of iron (II) and H⁺ will be = 0.022499 +  5.076*10⁻³ = 0.02758 M

So, pH = -log[H⁺]

pH = -log0.02758

pH = 1.56

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