Exercises — Water — structure (HOH = 104.5°), anomalous expansion, hardness (temporary - permanent), softening

The angle is $==104.5^{\circ }==$. The ideal tetrahedral angle is $109.5^{\circ }$. Water has two lone pairs on oxygen. Because a lone pair is held by only one nucleus (not shared between two), it spreads out and pushes harder than a bonding pair. Two such lone pairs squeeze the H–O–H angle down from $109.5^{\circ }$ to $104.5^{\circ }$.

Liquid water is densest at $==4^{\circ }\text{C}==$ ($\rho \approx 1.000{\text{ g/cm}}^3$). Ice is less dense ($\rho \approx 0.917{\text{ g/cm}}^3$) — that is why ice floats. See the density curve figure below.

Rule of thumb: if the salt contains ${\text{HCO}}_3^{-}$ (bicarbonate), it is temporary. Everything else is permanent.

Step 1 — WHAT reaction. On heating, bicarbonate decomposes: ${\text{Ca(HCO}}_3{)}_2\stackrel{\Delta }{\to }{\text{CaCO}}_3\downarrow +{\text{H}}_2\text{O}+{\text{CO}}_2\uparrow$ WHY: heat drives off ${\text{CO}}_2$, and the leftover ${\text{CO}}_3^{2-}$ instantly grabs ${\text{Ca}}^{2+}$ as insoluble ${\text{CaCO}}_3$ (this is Le Chatelier at work — removing a gas pulls the equilibrium forward).

Step 2 — mole ratio is $1:1$. So $n({\text{CaCO}}_3)=0.02\text{ mol}$.

Step 3 — mass. $M({\text{CaCO}}_3)=40+12+3(16)=100\text{ g/mol}$. $\text{mass}=0.02\times 100=\boxed{2.0\text{ g}}$

Step 1 — mass is conserved. Freezing does not create or destroy water, so $m$ is fixed. Step 2 — volume from $\rho =m/V\Rightarrow V=m/\rho$. The same mass at lower density needs more volume: $\frac{V_{\text{ice}}}{V_{\text{water}}}=\frac{{\rho }_{\text{water}}}{{\rho }_{\text{ice}}}=\frac{1.000}{0.917}=1.0905$ Step 3 — percent expansion. $(1.0905-1)\times 100=\boxed{\approx 9.1\%}$ This $\sim 9\%$ expansion is exactly why the sealed container bursts and why frozen pipes crack.

All three centres are surrounded by four electron pairs → all start from the tetrahedral $109.5^{\circ }$ template (VSEPR Theory). The difference is the count of lone pairs:

WHY lone pairs push harder: a lone pair is anchored to only one nucleus, so its electron cloud is fatter and closer to the central atom. It occupies more angular space and repels bonding pairs more strongly than a bonding pair does. Each added lone pair steals roughly $2$–$3^{\circ }$ from the bond angle. This is the charge-cloud geometry that also makes water polar.

Step 1 — count ends. Each molecule sits at the end of $4$ hydrogen bonds, giving $4N$ bond-ends. Step 2 — don't double count. Every hydrogen bond has two ends (one on each molecule it joins). So dividing by $2$ removes the double count: $\text{bonds}=\frac{4N}{2}=\boxed{2N}$ So $N$ water molecules give $2N$ hydrogen bonds — twice as many bonds as molecules. This dense, rigid cage is what makes ice both open (low density) and mechanically strong.

Step 1 — the reaction. Sulphate hardness cannot be boiled off, so we precipitate calcium as its insoluble carbonate: ${\text{CaSO}}_4+{\text{Na}}_2{\text{CO}}_3\to {\text{CaCO}}_3\downarrow +{\text{Na}}_2{\text{SO}}_4$ WHY it works: solubility rules say sulphates are soluble but ${\text{CaCO}}_3$ has a tiny $K_{sp}$ — so carbonate ion drags ${\text{Ca}}^{2+}$ out as solid.

Step 2 — total moles of ${\text{Ca}}^{2+}$. $n({\text{CaSO}}_4)=0.005\frac{\text{mol}}{\text{L}}\times 1000\text{ L}=5\text{ mol}$ Step 3 — $1:1$ ratio → $n({\text{Na}}_2{\text{CO}}_3)=5\text{ mol}$. Step 4 — mass. $\text{mass}=5\times 106=\boxed{530\text{ g}}$

WHY "CaCO₃ equivalent": water chemists report all hardness on one common scale — how much ${\text{CaCO}}_3$ would carry the same amount of charge (both ${\text{Ca}}^{2+}$ and ${\text{Mg}}^{2+}$ are $2+$, matching ${\text{CaCO}}_3$'s ${\text{Ca}}^{2+}$). So we convert each ion mass to "equivalent mass of ${\text{CaCO}}_3$".

Step 1 — convert Ca²⁺. $1{\text{ mol Ca}}^{2+}\equiv 1{\text{ mol CaCO}}_3$: $40\frac{\text{mg}}{\text{L}}\times \frac{100}{40}=100\text{ ppm}$ Step 2 — convert Mg²⁺. $12\frac{\text{mg}}{\text{L}}\times \frac{100}{24}=50\text{ ppm}$ Step 3 — add. $100+50=\boxed{150{\text{ ppm as CaCO}}_3}$ By convention this is "hard" water (over $120$ ppm).

Strategy — free first, then chemical. Boiling is essentially free, but only touches bicarbonates. So boil first, then dose the leftover permanent hardness with washing soda. Reversing the order would waste washing soda on the temporary part too.

(a) Stage assignment

• Boiling → removes the ${\text{Ca(HCO}}_3{)}_2$ (temporary).
• ${\text{Na}}_2{\text{CO}}_3$ → removes the ${\text{CaSO}}_4$ (permanent).

(b) Scale from boiling. $n({\text{Ca(HCO}}_3{)}_2)=0.003\times 500=1.5\text{ mol}$ $1:1$ → $1.5{\text{ mol CaCO}}_3$, and $M=100$: $\text{mass}=1.5\times 100=\boxed{150\text{ g scale}}$

(c) Washing soda for the sulphate. $n({\text{CaSO}}_4)=0.002\times 500=1.0\text{ mol}$ $1:1$ with ${\text{Na}}_2{\text{CO}}_3$ ($M=106$): $\text{mass}=1.0\times 106=\boxed{106{\text{ g Na}}_2{\text{CO}}_3}$ A modern plant would instead pass the water through ion-exchange resins, but the boil-then-dose route is the classic manual method.

Track density $\rho$ against temperature (figure s01) and remember denser water sinks.

1. Above $4^{\circ }\text{C}$ (normal regime): as surface water cools, it gets denser and sinks; warmer water rises to be cooled. The whole lake mixes and cools uniformly toward $4^{\circ }\text{C}$.
2. Exactly $4^{\circ }\text{C}$: density is at its maximum ($\rho \approx 1.000{\text{ g/cm}}^3$). This densest water settles at the bottom. The whole lake is now $\ge 4^{\circ }\text{C}$ with the coldest-densest at the base.
3. $4^{\circ }\text{C}\to 0^{\circ }\text{C}$ (anomalous regime): further cooling of the surface now makes it less dense (hydrogen bonds start forming open cages). So this colder water stays on top — it no longer sinks. The lake stratifies: $\approx 4^{\circ }\text{C}$ below, colder above.
4. At $0^{\circ }\text{C}$ (freezing): the topmost layer freezes into ice ($\rho =0.917$), which is lighter still and floats. This ice sheet insulates the water beneath.

Conclusion: because the densest water is $4^{\circ }\text{C}$ (not $0^{\circ }\text{C}$), lakes freeze top-down, leaving a liquid $\sim 4^{\circ }\text{C}$ layer at the bottom. Fish and aquatic life survive there — a direct consequence of water's anomalous expansion, and a keystone fact for Aquatic Ecosystems.

Water bond angle ::: $104.5^{\circ }$, reduced from $109.5^{\circ }$ by two lone pairs Temperature of max liquid density ::: $4^{\circ }\text{C}$ Percent water expands on freezing ::: about $9\%$ Test to spot temporary hardness ::: it disappears on boiling (forms ${\text{CaCO}}_3$ scale) Reagent for permanent hardness ::: washing soda, ${\text{Na}}_2{\text{CO}}_3$ Hydrogen bonds per $N$ molecules in ice ::: $2N$ Conversion factor to ${\text{CaCO}}_3$ equivalent ::: $M_{{\text{CaCO}}_3}÷M_{\text{ion}}$