Chapter 15, Problem 50QAP

Interpretation Introduction

(a)

Interpretation:

To calculate K for the following reaction if for $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}{\text{K}}_{\text{sp}}{\text{=2×10}}^{\text{-19}}$ and for $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}^{\text{2+}}{\text{K}}_{\text{f}}{\text{=2×10}}^{\text{12}}$ :

$\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right){\text{+4NH}}_{\text{3}}\left(\text{aq}\right)\stackrel{}{\rightleftarrows }\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}^{\text{2+}}\left(\text{aq}\right){\text{+2OH}}^{\text{-}}\left(\text{aq}\right)$

Concept introduction:

Equilibrium constant (k) is the expression of reversible reaction which shows the relationship between the presence of amount of reactant and product.

$\begin{array}{l}\text{A+B}\rightleftharpoons \text{C+D}\\ K=\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]\left[\text{B}\right]}\end{array}$

Here, K is equilibrium constant, $\left[\text{A}\right]$, $\left[\text{B}\right]$$\left[\text{C}\right]$$\left[\text{D}\right]$ are concentration of A, B, C and D

Interpretation Introduction

(b)

Interpretation:

To determine the solubility of $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(\text{in}\text{mol/L}\right)$ in $\text{4}\text{.5}\text{M}{\text{NH}}_{\text{3}}\text{.}$

Concept introduction:

The number of solutes that can be dissolved in one liter solution before the solution becomes saturated is termed as molar solubility.