Answered: Calculate the molar solubility of…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Calculate the molar solubility of Fe(OH)2 in pure water

Expert Solution

Step 1

The molar solubility has a direct relation with the solubility product. The solubility product of a substance is the number of moles of solute dissolved in per liter volume of the solution until the solution gets saturated ( means the maximum amount of solute is dissolved). Its unit of measurement is mol/l or M (molarity). For example,

$A_{a} B_{b} (s) \leftrightarrow {aA}^{+} (aq) + {bB}^{-} (aq) K_{sp} = {[A^{+}]}^{a} {[B^{-}]}^{b} where, K_{sp} is solubility product constant A^{+} is cation in aqueous solution B^{-} is anion in aqueous solution a and b are relative concentrations of A and B$

Step 2

The reaction of solubility of Fe(OH)₂ in pure water is as-

$Fe (OH)_{2} (s) \leftrightarrow {Fe}^{2 +} (aq) + 2 {OH}^{-} (aq)$

Assume x be the molar solubility of Fe(OH)₂which gives x mol/l of Fe²⁺ and 2x mol/l of OH^-.

The K_sp value of Fe(OH)₂is $4.87 \times 10^{- 17}$

Now, writing the solubility product relation with molar solubility for the above equation is as-

$\begin{array}{rcl} K_{s p} & = & [F e^{2 +}] {[O H^{-}]}^{2} 4.87 \times 10^{- 17} = x . {(2 x)}^{2} 4.87 \times 10^{- 17} = 4 x^{3} \\ x & = & \frac{\sqrt[3]{4.87 \times 10^{- 17}}}{4} \\ = & 2.30 \times 10^{- 6} M \end{array}$

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