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

This page is the practice arena for the Water topic note. Before we solve anything, we lay out every kind of situation a problem can throw at you. Then each worked example is tagged with which situation ("cell") it covers, so by the end no scenario is left unseen.

Everything here rests on ideas built in the parent note. Where a tool enters, we say why that tool and not another.

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The scenario matrix

Think of hardness/water problems as living in a grid. The columns are the kind of quantity asked; the rows are the kind of input given. Our job in D3 is to touch every cell at least once.

Cell	Scenario class	What makes it tricky	Covered by
A	Temporary hardness, boiling, 1:1 stoichiometry	trust the mole ratio	Ex 1
B	Permanent hardness, washing-soda dose	two different salts, mixed molar masses	Ex 2
C	Mixed hardness (temp + perm together)	must split the total	Ex 3
D	Zero / degenerate input (already soft water)	the "nothing happens" case	Ex 4
E	Hardness expressed as ppm CaCO₃ (the standard unit)	unit conversion, why we use CaCO₃	Ex 5
F	Real-world word problem (kettle scale over a month)	scaling up, sanity of magnitude	Ex 6
G	Structure/geometry — bond angle & dipole reasoning	no numbers to "plug", reason from VSEPR	Ex 7
H	Anomalous expansion — density limiting behaviour	the 4 °C turning point, sign of change	Ex 8
I	Exam-style twist — soap wasted before lather	link ions → precipitate → threshold	Ex 9

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Example 1 — Cell A: Temporary hardness removed by boiling

Forecast: guess before reading — will the answer be near $1\text{g}$, $3\text{g}$, or $30\text{g}$?

Step 1 — Find total moles of the bicarbonate. $n=0.015\frac{\text{mol}}{\text{L}}\times 2\text{L}=0.030\text{mol}$ Why this step? Concentration alone can't give a mass; we need the actual amount, so we multiply by the volume that carries it.

Step 2 — Use the decomposition equation. $\text{Ca}({\text{HCO}}_3{)}_2\stackrel{\Delta }{\to }{\text{CaCO}}_3\downarrow +{\text{H}}_2\text{O}+{\text{CO}}_2\uparrow$ The ratio $\text{Ca}({\text{HCO}}_3{)}_2:{\text{CaCO}}_3$ is $1:1$, so $n_{{\text{CaCO}}_3}=0.030\text{mol}$. Why this step? One calcium atom goes in, one comes out as solid — no calcium is created or destroyed, so moles pass through unchanged (see Le Chatelier's Principle: heat drives the equilibrium toward the escaping ${\text{CO}}_2$).

Step 3 — Convert moles to grams. Molar mass $M({\text{CaCO}}_3)=40+12+3(16)=100\text{g/mol}$. $m=0.030\text{mol}\times 100\frac{\text{g}}{\text{mol}}=3.0\text{g}$

Verify: Units: $\text{mol}\times \text{g/mol}=\text{g}$ ✓. Magnitude: $0.03\text{mol}$ of a $100\text{g/mol}$ solid ≈ a few grams — consistent with visible kettle scale, not micrograms and not kilograms. ✓

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Example 2 — Cell B: Permanent hardness, washing-soda dose

Forecast: more or less than the $0.010\text{mol}$ of metal ions present?

Step 1 — Write both precipitation reactions. ${\text{CaSO}}_4+{\text{Na}}_2{\text{CO}}_3\to {\text{CaCO}}_3\downarrow +{\text{Na}}_2{\text{SO}}_4$ ${\text{MgCl}}_2+{\text{Na}}_2{\text{CO}}_3\to {\text{MgCO}}_3\downarrow +2\text{NaCl}$ Why this step? Solubility Rules tell us sulfates and chlorides are soluble but the carbonates of ${\text{Ca}}^{2+}/{\text{Mg}}^{2+}$ are not — so ${\text{CO}}_3^{2-}$ is the reagent that pulls the ions out.

Step 2 — Count carbonate needed. Each salt is $1:1$ with ${\text{Na}}_2{\text{CO}}_3$: $n({\text{Na}}_2{\text{CO}}_3)=0.004+0.006=0.010\text{mol}$ Why this step? Every divalent ion needs exactly one carbonate to become one neutral solid — charge balance, not mass, sets the ratio.

Step 3 — Convert to grams. $M({\text{Na}}_2{\text{CO}}_3)=2(23)+12+3(16)=106\text{g/mol}$. $m=0.010\text{mol}\times 106\frac{\text{g}}{\text{mol}}=1.06\text{g}$

Verify: Moles of soda equal total moles of hardness ions ($0.010$) — exactly the forecast tie. ✓ Units cancel to grams. ✓

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Example 3 — Cell C: Mixed hardness split into its parts

Forecast: all removed, none removed, or partly removed?

Step 1 — Turn concentrations into amounts in the 1 L sample. Because the sample volume is $V=1\text{L}$, multiply each concentration by it: $n_{\text{temporary}}=0.005\frac{\text{mol}}{\text{L}}\times 1\text{L}=0.005\text{mol}$ $n_{\text{total}}=0.008\frac{\text{mol}}{\text{L}}\times 1\text{L}=0.008\text{mol},n_{\text{permanent}}=0.008-0.005=0.003\text{mol}$ Why this step? "How many moles" is an amount, not a concentration — we must attach the litre before subtracting, or the units don't match. (Here $V=1\text{L}$ makes the numbers look the same, but the multiplication is what makes them legal.)

Step 2 — Apply boiling to the temporary part only. Boiling touches only the bicarbonate calcium. The $0.005\text{mol}$ of temporary calcium precipitates as ${\text{CaCO}}_3$ and leaves solution. Why this step? Heat decomposes ${\text{HCO}}_3^{-}$; ${\text{Cl}}^{-}$ is untouched, so the chloride calcium is unmoved.

Step 3 — What survives? $n({\text{Ca}}_{\text{left}}^{2+})=0.003\text{mol}$ Why this step? ${\text{CaCl}}_2$ is a soluble strong electrolyte — heat does nothing to it, so its ions stay.

Verify: Removed + remaining $=0.005+0.003=0.008\text{mol}$ = original total. ✓ Partly-removed matches the forecast for a mixed sample. ✓

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Example 4 — Cell D: Degenerate case (already soft water)

Forecast: zero, or a trace?

Step 1 — Check the inputs. No divalent hardness ions present, so $n(\text{ions})=0\text{mol}$. Why this step? Every softening reaction consumes a metal ion; with none present, both reactions have nothing to react with.

Step 2 — Compute the amount of each reagent/product, then convert to mass. Moles first, molar mass second: $n({\text{Na}}_2{\text{CO}}_3)=0\text{mol}\Rightarrow m=0\text{mol}\times 106\frac{\text{g}}{\text{mol}}=0\text{g}$ $n({\text{CaCO}}_3)=0\text{mol}\Rightarrow m=0\text{mol}\times 100\frac{\text{g}}{\text{mol}}=0\text{g}$ Why this step? Even when the answer is zero, moles and grams are different quantities — you only reach grams after multiplying moles by the molar mass. Skipping that link is the habit that causes unit-slips on non-zero problems.

Verify: The "nothing happens" branch is real and legal — never assume a positive answer. Soft water needs no treatment. ✓ This is the degenerate cell that many students forget exists.

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Example 5 — Cell E: Hardness as ppm CaCO₃

Forecast: tens, hundreds, or thousands of ppm?

Step 1 — Map each ${\text{Mg}}^{2+}$ to one "equivalent" ${\text{CaCO}}_3$. $n({\text{CaCO}}_3\text{equiv})=n({\text{Mg}}^{2+})=0.001\text{mol/L}$ Why this step? One divalent ion equals the trouble of one ${\text{CaCO}}_3$ formula unit — that is the definition of the equivalence yardstick (see the intuition box above).

Step 2 — Convert to mass per litre. $0.001\frac{\text{mol}}{\text{L}}\times 100\frac{\text{g}}{\text{mol}}=0.1\text{g/L}=100\text{mg/L}$

Step 3 — Read off ppm. $100\text{mg/L}=100\text{ppm as }{\text{CaCO}}_3$

Verify: $100\text{ppm}$ falls in the "moderately hard" band ($60$–$120\text{ppm}$) used by water boards — a physically sensible number. ✓ Hundreds, as the higher forecast guessed. ✓

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Example 6 — Cell F: Real-world kettle scale over a month

Forecast: grams, tens of grams, or a kilogram?

Step 1 — Moles of bicarbonate boiled per day. $0.002\frac{\text{mol}}{\text{L}}\times 3\text{L}=0.006\text{mol/day}$ Why this step? We scale concentration by the daily volume first, because the deposit accumulates day by day.

Step 2 — Convert daily moles to daily mass of ${\text{CaCO}}_3$ (1:1, $M=100\text{g/mol}$): $0.006\text{mol}\times 100\frac{\text{g}}{\text{mol}}=0.6\text{g/day}$

Step 3 — Multiply over the month. $0.6\text{g/day}\times 30\text{day}=18\text{g}$

Verify: $18\text{g}$ of chalky crust over a month is realistic for a hard-water kettle — not a speck, not a brick. ✓ Tens of grams, as forecast. ✓

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Example 7 — Cell G: Structure & geometry (no plug-in numbers)

Forecast: will removing lone pairs widen or narrow the angle?

Figure 1 — how to read it: the blue sphere is the oxygen atom, the two white spheres are hydrogens joined by red O–H bonds, the two thick yellow lobes above are the lone pairs, and the green arc marks the $104.5^{\circ }$ H–O–H angle. The green vertical arrow is the net dipole pointing from the H side up toward O. The $x$–$y$ axes are just a drawing plane (positions in arbitrary units); notice how the yellow lobes press down on the red bonds, squeezing them together.

Step 1 — Count electron pairs on the central atom. Oxygen: 2 bonding pairs (to H) + 2 lone pairs = 4 pairs total → tetrahedral electron geometry (~$109.5^{\circ }$), exactly as VSEPR Theory predicts. Why this step? The number of pairs sets the base shape before any distortion.

Step 2 — Apply the repulsion ranking. $\text{LP–LP}>\text{LP–BP}>\text{BP–BP}$ Two lone pairs, held by a single nucleus, spread wider and squeeze the two O–H bonds together, dropping the angle from $109.5^{\circ }$ to $104.5^{\circ }$ (look at the yellow lone-pair lobes crowding the red bonds in the figure). Why this step? Lone pairs are "fatter" clouds; the bonded pairs yield to them.

Step 3 — Get polarity from the bent shape. Because the molecule is bent (not linear), the two O–H bond dipoles do not cancel; they add to a net dipole (the green arrow) pointing from the H side toward O ($\mu =1.85\text{D}$). This is why water hydrogen-bonds — see Electronegativity and Polarity and Hydrogen Bonding.

Verify: Trend check — ${\text{CH}}_4(0\text{LP})=109.5^{\circ }$, ${\text{NH}}_3(1\text{LP})=107^{\circ }$, ${\text{H}}_2\text{O}(2\text{LP})=104.5^{\circ }$: each extra lone pair narrows the angle, and both drops sit in the $2$–$3^{\circ }$ range. ✓ Narrower, as forecast.

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Example 8 — Cell H: Anomalous expansion, limiting behaviour

Forecast: does water simply expand throughout, or reverse somewhere?

Figure 2 — how to read it: the horizontal axis is temperature in ${}^{\circ }\text{C}$ ($0$ to $10$); the vertical axis is density in ${\text{g/cm}}^3$. The blue curve rises then falls, peaking at the yellow dot at $4^{\circ }\text{C}$ (maximum density $\approx 0.99997{\text{g/cm}}^3$). The green arrow on the left marks the "cages collapse → density up" regime below $4^{\circ }\text{C}$; the red arrow on the right marks the "thermal expansion → density down" regime above it. The white dot at $0^{\circ }\text{C}$ is liquid water just above freezing.

Step 1 — Split the range at the turning point.

• $0^{\circ }\text{C}\to 4^{\circ }\text{C}$: open hydrogen-bond cages collapse → volume decreases → density increases.
• $4^{\circ }\text{C}\to 10^{\circ }\text{C}$: ordinary thermal motion wins → volume increases → density decreases.

Why this step? Two opposing effects (cage-breaking vs. thermal expansion) trade dominance; the answer needs both branches, not one.

Step 2 — Locate the extremum. The two effects exactly balance at $4^{\circ }\text{C}$, giving maximum density $\approx 0.99997{\text{g/cm}}^3$ (the peak of the curve in the figure).

Step 3 — Limiting/degenerate check at $0^{\circ }\text{C}$. Liquid water at $0^{\circ }\text{C}$ has density $\approx 0.9998{\text{g/cm}}^3$ — slightly less than the $4^{\circ }\text{C}$ peak. On freezing to ice the density drops sharply to $0.917{\text{g/cm}}^3$ — a $\sim 9\%$ increase in volume, so ice floats.

Verify: Density at $0^{\circ }\text{C}$ water ($0.9998$) $<$ density at $4^{\circ }\text{C}$ ($0.99997$) $>$ density at $10^{\circ }\text{C}$ ($0.9997$): a genuine maximum, not a monotonic curve. ✓ Important for Aquatic Ecosystems — surface ice insulates life below.

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Example 9 — Cell I: Exam twist, soap wasted before lather

Forecast: equal to the ${\text{Ca}}^{2+}$ moles, double, or half?

Step 1 — Write the scum reaction. $2{\text{C}}_{17}{\text{H}}_{35}{\text{COO}}^{-}+{\text{Ca}}^{2+}\to ({\text{C}}_{17}{\text{H}}_{35}\text{COO}{)}_2\text{Ca}\downarrow$ Why this step? Lather is only possible once every ${\text{Ca}}^{2+}$ is precipitated (see Soap Chemistry); until then soap is consumed, not cleaning.

Step 2 — Apply the $2:1$ ratio. Each ${\text{Ca}}^{2+}$ eats two stearate ions: $n_{\text{soap wasted}}=2\times 0.0005=0.0010\text{mol}$ Why this step? Calcium is $2+$, stearate is $1-$; charge balance forces two soap anions per calcium — the twist that catches students who guess $1:1$.

Step 3 — Interpret. Only after $0.0010\text{mol}$ of soap is lost does the next soap added begin to lather.

Verify: Ratio $=2$, matching the "double" forecast. ✓ Units: dimensionless mole ratio times moles gives moles. ✓ (Ion-exchange softening via Ion Exchange Resins would avoid this waste entirely by removing ${\text{Ca}}^{2+}$ first.)

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Recall Self-test before you leave

Which cell forces a $2:1$ ratio, and why? ::: Cell I (soap/scum) — because ${\text{Ca}}^{2+}$ is divalent and stearate is monovalent, so two anions per calcium. Washing soda needed for $0.010\text{mol}$ total divalent hardness? ::: $0.010\text{mol}\times 106\text{g/mol}=1.06\text{g}$ of ${\text{Na}}_2{\text{CO}}_3$ (1:1 per ion). Water's density maximum sits at which temperature, and why there? ::: At $4^{\circ }\text{C}$, where cage-collapse (below) and thermal expansion (above) exactly balance. Boiling removes which hardness? ::: Only temporary (bicarbonate) hardness; permanent (sulfate/chloride) survives.