Question

Above what fe2 concentration will fe(oh)2 precipitate from a buffer solution that has a ph of 8.18? the ksp of fe(oh)2 is 4.87×10-17.

Answer 1

The solubility equilibrium is represented by the following reaction:

Fe(OH)2 (s) ↔ Fe2+(aq) + 2OH-(aq)

Ksp = [Fe2+][OH-]²------(1)

It is given that: pH = 8.18

Now, pH + pOH = 14

pOH = 14 - 8.18 = 5.82

[OH-] = 10^-pOH = 10^-5.85 = 1.51*10⁻⁶M

From eq(1)

[Fe2+] = Ksp/[OH-]² = 4.87*10⁻¹⁷/(1.51*10⁻⁶)² = 2.14*10⁻⁵M

Ans: [Fe2+] = 2.14*10⁻⁵M

Answer 2

The concentration of the Fe²⁺ that will precipitate from the buffer solution that has a pH of 8.18 is 2.14×10¯⁵ M

How to determine the pOH of the solution

• pH = 8.18
• pOH =?

pH + pOH = 14

8.18 + pOH = 14

Collect like terms

pOH = 14 – 8.18

pOH = 5.82

How to determine the concentration of OH¯

• pOH = 5.82
• Concentration of OH¯ =?

pOH = –Log [OH¯]

5.82 = –Log [OH¯]

Multiply through by –1

–5.82 = Log [OH¯]

Take the anti-log of –5.82

[OH¯] = anti-log ( –5.82)

[OH¯] = 1.51×10⁻⁶ M

How to determine the concentration of Fe²⁺

Fe(OH)₂ (aq) <=> Fe²⁺(aq) + 2OH¯(aq)

• Solubility of product (Ksp) = 4.87×10¯¹⁷
• Concentration of [OH¯] = 1.51×10⁻⁶ M

• Concentration of [Fe²⁺] = ?

Ksp = [Fe²⁺] [OH¯]²

[Fe²⁺] = Ksp / [OH¯]²

[Fe²⁺] = 4.87×10¯¹⁷ / (1.51×10⁻⁶)²

[Fe²⁺] = 2.14×10¯⁵ M

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