Chapter 15, Problem 15.139CHP

(a)

Interpretation Introduction

Interpretation:

Using the equilibrium constant and ${\text{K}}_{\text{f}}$ value for $\text{Fe}\left(\text{Cit}\right)$, the equilibrium constant K for given reaction has to be calculate.

Concept Introduction:

Equilibrium constant (K):

In the equilibrium reaction, the ratio of concentration of the reactant and concentration of the product.

Reaction quotient (Q): It is defined as the ratio between concentration of the product and concentration of the reactant.

$\text{Q =}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$

This value used to know about the direction of the reaction is going to proceed.

If $\text{Q = K}$ means the reaction reached the equilibrium.

If $\text{Q > K}$ means, the reaction must go to left.

If $\text{Q < K}$ means, the reaction must go to right.

(b)

Interpretation Introduction

Interpretation:

Molar solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ in $\text{0}\text{.500}\text{M}{\text{H}}_{\text{3}}\text{Cit}$ has to be calculated.

Concept Introduction:

Molar solubility: The number of moles of a solute dissolved in litre of the solution before the solution is saturated.  It can be measured by the solubility product constant $\left({\text{K}}_{\text{sp}}\right)$ of the substance and stoichiometry.

The general reaction is,

${\text{M}}_{\text{n}}{\text{X}}_{\text{m}\left(\text{s}\right)}\stackrel{}{\leftrightarrow }{\text{ nM }}_{\left(\text{aq}\right)}^{\text{m+}}{\text{ + mX }}_{\left(\text{aq}\right)}^{\text{n-}}$

$\text{Molar solubility = }\sqrt[\left(\text{n+m}\right)]{\frac{{\text{K}}_{\text{sp}}}{{\text{n}}^{\text{n}}·{\text{m}}^{\text{m}}}}$