Question

Calculate the solubility of CaF2 in g/L (Ksp = 4.0 x 10-8). 2. What is the pH of a solution containing a hydrogen ion concentration of 3.0x10-4M? 3. What is the oxidizing agent and reducing agent in the following reaction: Zn(s) + Cl2(g) → ZnCl2(s) 4. Assign the oxidation numbers of the atoms in the following compounds. (a) Al2O3 (b) XeF4 (c) K2Cr2O7

Answer 1

Answer:

$\large \boxed{1. \text{ 0.17 g/L; 2. 3.52; 3. Cl; 4. (a) +3; (b) +4; (c) +6}}$

Explanation:

1. Solubility of CaF_2

(a) Molar solubility

CaF₂ ⇌ Ca²⁺ + 2F⁻

$K_{\text{sp }} = \text{[Ca ^{2+} ]}\text{[F ^{-} ]}^{2}= 4.0 \times 10^{-8}\\s(2s)^{2}=4.0 \times 10^{-8}\\4s^{3} = 4.0 \times 10^{-8}\\s^{3} = 1.0 \times 10^{-8}\\s =2.2 \times 10^{-3}\text{ mol/L}$

(b) Mass solubility

$\text{Solubility} = 2.2 \times 10^{-3} \text{ mol/L} \times \dfrac{\text{78.07 g}}{\text{1 L }} = \text{0.17 g/L}\\\\\text{The solubility of CaF _{2} is \large \boxed{\textbf{0.17 g/L}} }$

2. pH

pH = -log [H⁺] = -log(3.0 × 10⁻⁴) = 3.52

3. Oxidizing and reducing agents

Zn + Cl₂ ⟶ ZnCl₂

$\rm \stackrel{\hbox{0}}{\hbox{Zn}} + \stackrel{\hbox{0}}{\hbox{ Cl}_{2} }\longrightarrow \stackrel{\hbox{+2}}{\hbox{Zn}}\stackrel{\hbox{-1}}{\hbox{Cl}_{2}}$

The oxidation number of Cl has decreased from 0 to -1.

Cl has been reduced, so Cl is the oxidizing agent.

4. Oxidation numbers

(a) Al₂O₃

$\stackrel{\hbox{ \mathbf{+3} }}{\hbox{Al}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{3}}$

1O = -2; 3O = -6; 2Al  = +6; 1Al = +3

(b) XeF₄

$\stackrel{\hbox{ \mathbf{+4} }}{\hbox{Xe}}\stackrel{\hbox{-1}}{\hbox{F}_{4}}$

1F = -1; 4F = -4; 1 Xe = +4

(c) K₂Cr₂O₇

$\stackrel{\hbox{ {+1} }}{\hbox{K}_{2}}\stackrel{\hbox{ \mathbf{+6} }}{\hbox{Cr}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{7}}$

1K = +1; 2K = +2; 1O = -2; 7O = -14

+2 - 14 = -12

2Cr = + 12; 1 Cr = +6