Why can dilute solutions of hydrogen peroxide not be concentrated by heating? How can a concentrated solution of hydrogen peroxide be obtained?
Dilute solutions of $\mathrm{H}_2 \mathrm{O}_2$ cannot be concentrated by heating because it decomposes much below its boiling point.
$$2 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2$$
$1 \% \mathrm{H}_2 \mathrm{O}_2$ is extracted with water and concentrated to $\sim 30 \%$ (by mass) by distillation under reduced pressure. It can be further concentrated to $\sim 85 \%$ by careful distillation under low pressure. The remaining water can be frozen out to obtain pure $\mathrm{H}_2 \mathrm{O}_2$.
Why is hydrogen peroxide stored in wax lined bottles?
Hydrogen peroxide is decomposed by rough surfaces of glass, alkali oxides present in it and light to form $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{O}_2$.
$$2 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2$$
To prevent this decomposition, hydrogen peroxide is usually stored in paraffin wax coated plastic or teflon bottles.
Why does hard water not form lather with soap?
Hard water contains salts of calcium and magnesium ions. Hard water does not give lather with soap and forms scum/precipitate with soap. Soap containing sodium stearate $\left(\mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COONa}\right)$ reacts with hard water to precipitate out as $\mathrm{Ca} / \mathrm{Mg}$ stearate.
$$2 \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COONa}(a q)+\mathrm{M}^{2+}(\mathrm{aq}) \longrightarrow\left(\mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COO}\right)_2 \mathrm{M} \downarrow+2 \mathrm{Na}^{+}(a q)$$ (where, $M$ is $\mathrm{Ca} / \mathrm{Mg}$ )
It is therefore, unsuitable for laundry.
Phosphoric acid is preferred over sulphuric acid in preparing hydrogen peroxide from peroxides. Why?
$\mathrm{H}_2 \mathrm{SO}_4$ acts as a catalyst for decomposition of $\mathrm{H}_2 \mathrm{O}_2$. Therefore, some weaker acids such as $\mathrm{H}_3 \mathrm{PO}_4, \mathrm{H}_2 \mathrm{CO}_3$ is preferred over $\mathrm{H}_2 \mathrm{SO}_4$ for preparing $\mathrm{H}_2 \mathrm{O}_2$ from peroxides.
$$3 \mathrm{BaO}_2+2 \mathrm{H}_3 \mathrm{PO}_4 \longrightarrow \underset{\text { (insoluble) }}{\mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2}+3 \mathrm{H}_2 \mathrm{O}_2 $$
How will you account for $104.5^{\circ}$ bond angle in water?
In water, oxygen has $s p^3$-hybridisation and the bond angle of HOH should have been $109^{\circ} 28^{\prime}$. In $\mathrm{H}_2 \mathrm{O}$, the oxygen atom is surrounded by two shared pairs and two lone pairs of electrons. From VSEPR theory, lone pair - lone pair repulsions are stronger than bond pair-bond pair repulsions.
As a result, the bond angle of HOH in water slightly decreases from the regular tetrahedral angle of $109^{\circ} .28^{\prime}$ to $104.5^{\circ}$.
