Exercises — Solution stoichiometry — titrations, dilutions

WHAT tool? We want moles from a known concentration and volume → the master relation $n=cV$. Convert volume: $250\text{ mL}=250/1000=0.250\text{ L}$. Why? $c$ is per litre, so a raw mole count needs $V$ in litres. $n=cV=0.400\times 0.250=0.100\text{ mol}$ Answer: $0.100\text{ mol}$ of glucose. Sanity: a quarter-litre of a "point-four" solution — roughly a tenth of a mole. Feels right.

WHAT tool? Given moles and volume, find concentration → rearrange $n=cV$ to $c=n/V$. Convert volume: $200\text{ mL}=200/1000=0.200\text{ L}$. Why? We divide moles by a litre-volume to get ${\text{mol L}}^{-1}$. $c=\frac{n}{V}=\frac{0.0500}{0.200}=0.250\text{ M}$ Answer: $0.250\text{ M}$.

WHAT tool? Adding water → moles conserved → dilution equation $c_1{V}_1=c_2{V}_2$. Volumes appear on both sides, so I keep mL (units cancel). $c_2=\frac{c_1{V}_1}{V_2}=\frac{1.20\times 100}{600}=0.200\text{ M}$ Answer: $0.200\text{ M}$. Forecast-then-verify: volume grew $\times 6$, so concentration should fall $\times 6$: $1.20/6=0.200$. ✓

WHAT tool? Diluting concentrated stock → $c_1{V}_1=c_2{V}_2$, solve for $V_1$ (stock volume). Volumes both sides → keep mL. $V_1=\frac{c_2{V}_2}{c_1}=\frac{0.150\times 500}{2.00}=37.5\text{ mL}$ Water to add = final − stock = $500-37.5=462.5\text{ mL}$. Answer: take $37.5\text{ mL}$ of stock, add $462.5\text{ mL}$ water (top up to the $500\text{ mL}$ mark). Why "top up to the mark," not "add exactly 462.5"? Because molarity uses the total final volume; in practice you add water until the meniscus reaches the $500\text{ mL}$ line. The $462.5$ is a good estimate of how much water that will be.

WHAT tool? A titration with mole ratio $1:1$, so $\frac{c_A{V}_A}{1}=\frac{c_B{V}_B}{1}$, i.e. $c_A{V}_A=c_B{V}_B$. Volumes both sides → keep mL. $c_{\text{NaOH}}=\frac{c_{\text{HCl}}{V}_{\text{HCl}}}{V_{\text{NaOH}}}=\frac{0.100\times 18.0}{25.0}=0.0720\text{ M}$ Answer: $0.0720\text{ M}$. Sanity: less HCl volume than NaOH volume, at equal 1:1 ratio, means NaOH is more dilute than HCl — and indeed $0.072<0.100$. ✓

Step 1 — moles needed. Convert volume: $250\text{ mL}=250/1000=0.250\text{ L}$ (needed because $c$ is per litre). Then $n=cV=0.100\times 0.250=0.0250\text{ mol}$. Step 2 — mass. Using the molar-mass relation $n=m/M_r$ (see the $M_r$ definition above), rearrange to $m=n\times M_r$. $m=0.0250\times 106=2.65\text{ g}$ Answer: $2.65\text{ g}$. Weigh this precisely, dissolve, and top up to $250\text{ mL}$ — this is how a standard solution is made.

WHAT tool? Titration with ratio $1:2$ (acid : base). Use $\frac{c_A{V}_A}{a}=\frac{c_B{V}_B}{b}$ with $a=1$ (acid), $b=2$ (base). Step 1 — moles of base. Convert volume: $25.0\text{ mL}=25.0/1000=0.0250\text{ L}$ (needed because we want a raw mole count and $c$ is per litre). Then $n_{\text{KOH}}=cV=0.200\times 0.0250=5.00\times 10^{-3}\text{ mol}$. Step 2 — cross the ratio bridge. Each ${\text{H}}_2{\text{SO}}_4$ needs 2 KOH, so $n_{\text{acid}}=\frac{1}{2}{n}_{\text{KOH}}=\frac{1}{2}(5.00\times 10^{-3})=2.50\times 10^{-3}\text{ mol}$ Step 3 — concentration. Convert the acid volume: $20.0\text{ mL}=20.0/1000=0.0200\text{ L}$. Then $c=n/V=\frac{2.50\times 10^{-3}}{0.0200}=0.125\text{ M}$. Answer: $0.125\text{ M}$. Forecast-then-verify: if you forgot the 2, you'd get $0.250\text{ M}$ — exactly double the truth.

WHAT tool? Titration, ratio $1:2$ (base : acid), but now the unknown is a volume. Step 1 — moles of base. Convert volume: $30.0\text{ mL}=30.0/1000=0.0300\text{ L}$ (needed because we want a raw mole count from $c$ in per-litre). Then $n_{{\text{Ba(OH)}}_2}=cV=0.150\times 0.0300=4.50\times 10^{-3}\text{ mol}$. Step 2 — ratio bridge. Each ${\text{Ba(OH)}}_2$ needs 2 HCl (it has two OH⁻ to neutralise): $n_{\text{HCl}}=2\times 4.50\times 10^{-3}=9.00\times 10^{-3}\text{ mol}$ Step 3 — volume of HCl. $V=n/c=\frac{9.00\times 10^{-3}}{0.300}=0.0300\text{ L}$. Convert back to mL: $0.0300\text{ L}\times 1000=30.0\text{ mL}$. Answer: $30.0\text{ mL}$ of HCl. Why does it come out equal to the base volume? Coincidence of numbers: base is half as concentrated but each mole needs twice the acid — the factors cancel here. Don't rely on that; always run the ratio.

Stage A — find the diluted concentration. $c_1{V}_1=c_2{V}_2$ (volumes both sides → keep mL): $c_2=\frac{0.500\times 10.0}{250}=0.0200\text{ M NaOH}$ Stage B — moles of NaOH taken for titration. Convert volume: $20.0\text{ mL}=20.0/1000=0.0200\text{ L}$ (needed for a raw mole count). Then $n=cV=0.0200\times 0.0200=4.00\times 10^{-4}\text{ mol}$. Stage C — ratio bridge (1:1) → moles of HCl. $n_{\text{HCl}}=4.00\times 10^{-4}\text{ mol}$. Stage D — volume of HCl. $V=n/c=\frac{4.00\times 10^{-4}}{0.0400}=0.0100\text{ L}$. Convert back: $0.0100\text{ L}\times 1000=10.0\text{ mL}$. Answer: $10.0\text{ mL}$ of HCl. Workflow check: we did $c_1{V}_1=c_2{V}_2$ → $n=cV$ → ratio → $V=n/c$. That is "dilution, then CoVo → n → Ratio → n → CoVo." ✓

The idea: we can't easily titrate a slow-dissolving solid, so we drown it in a known amount of acid, then measure how much acid survived. $n_{\text{reacted with solid}}=n_{\text{added}}-n_{\text{leftover}}$ (see figure).

Step 1 — total HCl added. Convert volume: $50.0\text{ mL}=50.0/1000=0.0500\text{ L}$ (needed for a raw mole count). Then $n=cV=0.500\times 0.0500=0.0250\text{ mol}$. Step 2 — leftover HCl (found via the NaOH titration, ratio 1:1). Convert volume: $12.0\text{ mL}=12.0/1000=0.0120\text{ L}$. Then $n_{\text{leftover HCl}}=n_{\text{NaOH}}=0.250\times 0.0120=3.00\times 10^{-3}\text{ mol}$ Step 3 — HCl that reacted with the carbonate. $n_{\text{HCl reacted}}=0.0250-0.00300=0.0220\text{ mol}$ Step 4 — ratio bridge to ${\text{CaCO}}_3$ (ratio $1:2$, one carbonate per 2 HCl). $n_{{\text{CaCO}}_3}=\frac{1}{2}(0.0220)=0.0110\text{ mol}$ Step 5 — mass and purity. Using $m=n\times M_r$: $m=0.0110\times 100=1.10\text{ g}$ Sanity check against the sample mass: the sample weighs $1.50\text{ g}$ and we found $1.10\text{ g}$ of ${\text{CaCO}}_3$ — less than the sample, so this is physically possible. ✓ (Always run this check: if the pure mass exceeded the sample, we'd know a ratio was wrong.) Purity: $\text{purity}=\frac{1.10}{1.50}\times 100=73.3\%$ Answer: $1.10\text{ g}$ of ${\text{CaCO}}_3$, purity $=73.3\%$.

Design the chain: ${\text{KMnO}}_4$ (known) → moles ${\text{MnO}}_4^{-}$ → ratio bridge → moles ${\text{C}}_2{\text{O}}_4^{2-}$ → (1 oxalate per Ca) → moles ${\text{Ca}}^{2+}$ → divide by 1.00 L. Step 1 — moles of permanganate. Convert volume: $24.0\text{ mL}=24.0/1000=0.0240\text{ L}$ (needed for a raw mole count). Then $n_{{\text{MnO}}_4^{-}}=cV=0.0200\times 0.0240=4.80\times 10^{-4}\text{ mol}$. Step 2 — ratio bridge $2:5$. From the equation, $\frac{n_{{\text{MnO}}_4^{-}}}{2}=\frac{n_{\text{oxalate}}}{5}$, so $n_{\text{oxalate}}=\frac{5}{2}{n}_{{\text{MnO}}_4^{-}}=\frac{5}{2}(4.80\times 10^{-4})=1.20\times 10^{-3}\text{ mol}$ Step 3 — oxalate to calcium. In ${\text{CaC}}_2{\text{O}}_4$ there is one ${\text{Ca}}^{2+}$ per one ${\text{C}}_2{\text{O}}_4^{2-}$, so $n_{{\text{Ca}}^{2+}}=1.20\times 10^{-3}\text{ mol}$. Step 4 — concentration. The sample volume is already $1.00\text{ L}$, so $c=n/V=\frac{1.20\times 10^{-3}}{1.00}=1.20\times 10^{-3}{\text{ mol L}}^{-1}$. Answer: $n_{{\text{Ca}}^{2+}}=1.20\times 10^{-3}\text{ mol}$; concentration $=1.20\times 10^{-3}\text{ M}$.

Read the two stages carefully — this is a double-indicator titration. Stage 1 (to phenolphthalein end point, $15.0\text{ mL}$): only the carbonate reacts, converting to bicarbonate, 1:1. Convert volume: $15.0\text{ mL}=15.0/1000=0.0150\text{ L}$ (needed for a raw mole count). $n_{\text{HCl, stage 1}}=0.100\times 0.0150=1.50\times 10^{-3}\text{ mol}=n_{{\text{Na}}_2{\text{CO}}_3}$ Stage 2 (extra $40.0-15.0=25.0\text{ mL}$): this neutralises all bicarbonate present — both the original ${\text{NaHCO}}_3$ AND the bicarbonate just made from the carbonate. Convert volume: $25.0\text{ mL}=25.0/1000=0.0250\text{ L}$. $n_{\text{HCl, stage 2}}=0.100\times 0.0250=2.50\times 10^{-3}\text{ mol}=n_{{\text{HCO}}_3^{-}\text{ total}}$ The bicarbonate created from carbonate equals $n_{{\text{Na}}_2{\text{CO}}_3}=1.50\times 10^{-3}\text{ mol}$. So original bicarbonate: $n_{{\text{NaHCO}}_3}=2.50\times 10^{-3}-1.50\times 10^{-3}=1.00\times 10^{-3}\text{ mol}$ Concentrations. The solution volume is $25.0\text{ mL}=25.0/1000=0.0250\text{ L}$, so divide each mole count by $0.0250\text{ L}$: $c_{{\text{Na}}_2{\text{CO}}_3}=\frac{1.50\times 10^{-3}}{0.0250}=0.0600\text{ M},c_{{\text{NaHCO}}_3}=\frac{1.00\times 10^{-3}}{0.0250}=0.0400\text{ M}$ Answer: ${\text{Na}}_2{\text{CO}}_3=0.0600\text{ M}$, ${\text{NaHCO}}_3=0.0400\text{ M}$. Why subtract? Stage 2 titrates bicarbonate from two sources; only by removing the carbonate-derived part (known from Stage 1) do you isolate the original bicarbonate. Conservation of moles again.

"CoVo → n → Ratio → n → CoVo." Concentration × Volume gives moles; cross the mole-ratio bridge from the balanced equation; convert back to concentration or volume. Dilution is the special case where the "reaction" is just adding water and the ratio is $1:1$ with itself.