How unit 1 is examined
Hardness, its units and the EDTA method, alkalinity, and the numericals on them carry the marks; sources and impurities are short theory.
Sources
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Definition. Water sources are the natural stores from which raw water is drawn, and they are either surface water or underground water.
Key points.
- Rain water is the purest natural water, but it dissolves gases such as CO$_2$, SO$_2$ and dust while falling.
- River and lake water is surface water; it carries suspended clay, organic matter and microbes plus some dissolved salts.
- Well and spring water is underground water; it is clear and bacteria-poor but rich in dissolved Ca and Mg salts, so it is hard.
- Sea water is the most saline source (about 3.5% salts, mainly NaCl) and is unfit for drinking or boilers without treatment.
Impurities
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Definition. Impurities are the substances present in raw water besides H$_2$O, classified as suspended, dissolved, colloidal and biological.
Key points.
- Suspended impurities (sand, clay, silt, plankton) make water turbid and are removed by settling and filtration.
- Colloidal impurities (fine clay, silica, organic matter) do not settle and are removed by coagulation.
- Dissolved impurities are gases (O$_2$, CO$_2$) and salts (bicarbonates, chlorides, sulphates of Ca and Mg); the salts cause hardness.
- Biological impurities (bacteria, algae, fungi) are killed by disinfection.
Ozonation. Ozone is a strong oxidising disinfectant that decomposes as $\text{O}_3 \rightarrow \text{O}_2 + [\text{O}]$; the nascent oxygen destroys the cell walls of bacteria and pathogens and also removes colour and odour. It leaves no taste or odour and no residual chlorine by-products, but it is costly and gives no residual protection in the pipes.
Asked: [7 marks] (Nov 2022) How is portable water disinfected by ozonation?
Hardness and its units
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Definition. <mark>Hardness is the property of water by which it does not lather with soap, caused by dissolved bicarbonates, chlorides and sulphates of calcium and magnesium; it is expressed in mg/L (ppm) of CaCO$_3$ equivalent.</mark>
Key points.
- CaCO$_3$ is the standard because its molecular weight is 100 (equivalent weight 50), so the hardness of any salt converts to one common scale.
- Temporary (carbonate) hardness is caused by Ca and Mg bicarbonates and is removed by boiling.
- Permanent (non-carbonate) hardness is caused by chlorides and sulphates of Ca and Mg and is not removed by boiling.
- Total hardness = temporary + permanent hardness.
- Hard water wastes soap by forming insoluble scum and forms scale in boilers.
Formula. $\text{CaCO}_3\text{ equivalent} = \text{mass of salt} \times \dfrac{50}{\text{equivalent weight of salt}}$
| Unit | Meaning | Relation |
|---|---|---|
| ppm | 1 part CaCO$_3$ per $10^6$ parts water | 1 ppm = 1 mg/L |
| mg/L | mg of CaCO$_3$ per litre | 1 mg/L = 1 ppm |
| °Clark (°Cl) | 1 grain CaCO$_3$ per gallon (70,000 parts) | 1 °Cl = 14.3 ppm; 1 ppm = 0.07 °Cl |
| °French (°Fr) | 1 part CaCO$_3$ per $10^5$ parts | 1 °Fr = 10 ppm; 1 ppm = 0.1 °Fr |
Relation: 1 ppm = 1 mg/L = 0.07 °Cl = 0.1 °Fr.
| Basis | Temporary hardness | Permanent hardness |
|---|---|---|
| Cause | Bicarbonates of Ca and Mg | Chlorides and sulphates of Ca and Mg |
| Examples | Ca(HCO$_3$)$_2$, Mg(HCO$_3$)$_2$ | CaCl$_2$, MgCl$_2$, CaSO$_4$, MgSO$_4$ |
| Also called | Carbonate hardness | Non-carbonate hardness |
| Removal by boiling | Yes, bicarbonate precipitates as CaCO$_3$ / Mg(OH)$_2$ | No |
| Removal method | Boiling, Clark's (lime) process | Lime-soda, zeolite, ion exchange |
| Boiler effect | Soft sludge | Hard scale (CaSO$_4$) |
Answer frame. Open with the definition and the CaCO$_3$ standard; for units, give the four-unit table and the relation line; for the comparison, draw the six-row table with examples; close with total = temporary + permanent.
Asked: [7 marks] (Jun 2022) Explain various units of hardness of water giving their relationship. Asked: [7 marks] (Dec 2024) What is temporary and permanent hardness of water, how is it differentiated? Give suitable examples.
Determination of hardness by EDTA method
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Definition. EDTA (ethylenediaminetetraacetic acid, used as its disodium salt) is a complexing agent that forms stable 1:1 colourless complexes with Ca$^{2+}$ and Mg$^{2+}$.
Key points.
- Principle: the sample is buffered at pH 10 (NH$_4$Cl + NH$_4$OH) and titrated with EDTA using Eriochrome Black T (EBT).
- EBT first forms an unstable wine-red complex with Mg$^{2+}$/Ca$^{2+}$: $\text{M}^{2+} + \text{EBT} \rightarrow [\text{M-EBT}]$ (wine red).
- EDTA then takes the metal from EBT, forming a stable colourless complex: $[\text{M-EBT}] + \text{EDTA} \rightarrow [\text{M-EDTA}] + \text{EBT}$ (blue).
- End point: wine red to clear blue.
- Procedure: standardise EDTA with standard CaCO$_3$ (1 ml EDTA = $x$ mg CaCO$_3$); titrate 50 ml sample for total hardness; boil, filter and titrate for permanent hardness.
Formula. $\text{Hardness (ppm)} = \dfrac{V_{\text{EDTA}} \times x \times 1000}{V_{\text{sample}}}$
Asked: [7 marks] (Nov 2022) What is the principle of EDTA method? Explain the estimation of total hardness of water by complexometric method.
Alkalinity and its determination
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Definition. <mark>Alkalinity is the capacity of water to neutralise acid, caused by hydroxide (OH$^-$), carbonate (CO$_3^{2-}$) and bicarbonate (HCO$_3^-$) ions; it is expressed in ppm of CaCO$_3$.</mark>
Key points.
- Principle: a measured sample is titrated against standard acid (N/50 H$_2$SO$_4$) using two indicators in succession, phenolphthalein then methyl orange.
- Phenolphthalein end point (P): pink to colourless at pH about 8.3; here all OH$^-$ is neutralised and carbonate is only half neutralised: $\text{OH}^- + \text{H}^+ \rightarrow \text{H}_2\text{O}$; $\text{CO}_3^{2-} + \text{H}^+ \rightarrow \text{HCO}_3^-$.
- Methyl orange end point (M): yellow to red at pH about 4.5; the bicarbonate (original and formed) is neutralised: $\text{HCO}_3^- + \text{H}^+ \rightarrow \text{H}_2\text{O} + \text{CO}_2$.
- Phenolphthalein alkalinity P uses the acid up to the first end point; total alkalinity M uses the total acid up to the second end point.
- OH$^-$ and HCO$_3^-$ cannot coexist, because they react to give CO$_3^{2-}$; so only five combinations occur.
- Alkalinity above about 250 ppm is harmful in boilers, where it causes caustic embrittlement, priming and foaming.
Formula. $P = \dfrac{V_1 N \times 50 \times 1000}{V}$, $\quad M = \dfrac{(V_1+V_2) N \times 50 \times 1000}{V}$ ppm as CaCO$_3$
| Result | OH$^-$ | CO$_3^{2-}$ | HCO$_3^-$ |
|---|---|---|---|
| $P = 0$ | 0 | 0 | $M$ |
| $P = M$ | $M$ | 0 | 0 |
| $P = M/2$ | 0 | $2P$ | 0 |
| $P < M/2$ | 0 | $2P$ | $M - 2P$ |
| $P > M/2$ | $2P - M$ | $2(M-P)$ | 0 |
Answer frame. Open with the definition and the three ions; write the principle and both indicator reactions (points 2-3); draw the five-row table; close with the formulas for P and M and the note that OH$^-$ and HCO$_3^-$ never coexist. For the short note, give definition, ions, two indicators and the table in about half a page.
Asked: [7 marks] (Jun 2022) Discuss method for determination of alkalinity in given water sample. Asked: [14 marks] (Dec 2023, Dec 2024) Write brief note on (any two): i) Alkalinity (the other options belong to other units)
Related numerical problems
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Formula. $\text{CaCO}_3\text{ eq.} = \text{mass} \times \dfrac{50}{\text{Eq. wt}}$; Eq. wt = MW/2 for these salts: Ca(HCO$_3$)$_2$ 81, Mg(HCO$_3$)$_2$ 73, CaSO$_4$ 68, MgSO$_4$ 60, MgCl$_2$ 47.5, CaCl$_2$ 55.5.
Example 1 (Nov 2022, Jun 2025). Given Ca(HCO$_3$)$_2$ 4.86, Mg(HCO$_3$)$_2$ 5.84, CaSO$_4$ 6.8, MgSO$_4$ 8.4 mg/L.
| Salt | Working | CaCO$_3$ eq. (mg/L) |
|---|---|---|
| Ca(HCO$_3$)$_2$ | $4.86 \times 50/81$ | 3 |
| Mg(HCO$_3$)$_2$ | $5.84 \times 50/73$ | 4 |
| CaSO$_4$ | $6.8 \times 50/68$ | 5 |
| MgSO$_4$ | $8.4 \times 50/60$ | 7 |
Total hardness = 3 + 4 + 5 + 7 = 19 mg/L (ppm) as CaCO$_3$.
Example 2 (second part of the same question). Mg(HCO$_3$)$_2$ 7.3, Ca(HCO$_3$)$_2$ 16.4, MgCl$_2$ 9.5, CaSO$_4$ 13.6 mg/L. Temporary: $7.3 \times 50/73 = 5$ and $16.4 \times 50/81 = 10.12$ (10 if the paper reads 16.2). Permanent: $9.5 \times 50/47.5 = 10$ and $13.6 \times 50/68 = 10$. Temporary = 15.12, permanent = 20, total = 35.12 ppm.
Example 3 (Dec 2023, EDTA). Standard CaCO$_3$ = 15 g/L = 15 mg/ml, so 20 ml = 300 mg. 25 ml EDTA = 300 mg, so 1 ml EDTA = 12 mg CaCO$_3$.
- Total hardness = $18 \times 12 \times 1000/100$ = 2160 ppm.
- Permanent (boiled) = $12 \times 12 \times 1000/100$ = 1440 ppm.
- Temporary = 2160 - 1440 = 720 ppm.
Example 4 (Jun 2025, alkalinity). $P = 20 \times \tfrac{1}{50} \times 50 \times 1000/100 = 200$ ppm; $M = 22.5 \times \tfrac{1}{50} \times 50 \times 1000/100 = 225$ ppm. Since $P > M/2$ (112.5), OH$^-$ and CO$_3^{2-}$ are present: OH$^-$ = $2P - M$ = 175 ppm; CO$_3^{2-}$ = $2(M-P)$ = 50 ppm. Alkalinity is 175 ppm hydroxide + 50 ppm carbonate, total 225 ppm as CaCO$_3$.
Answer frame. Write Given, the formula, then one line per step, and end with the answer in bold with units.
Asked: [7 marks] (Nov 2022, Jun 2025) Calculate the total hardness of a water sample: Ca(HCO$_3$)$_2$ 4.86, Mg(HCO$_3$)$_2$ 5.84, CaSO$_4$ 6.8, MgSO$_4$ 8.4 mg/L; also temporary, permanent and total hardness for Mg(HCO$_3$)$_2$ 7.3, Ca(HCO$_3$)$_2$ 16.4, MgCl$_2$ 9.5, CaSO$_4$ 13.6 mg/L. Asked: [7 marks] (Dec 2023) 100 ml of water needed 18 ml EDTA; 20 ml standard CaCO$_3$ (15 g/L) needed 25 ml EDTA; the boiled sample needed 12 ml. Find temporary and permanent hardness. Asked: [7 marks] (Jun 2025) 100 ml water needed 20 ml N/50 H$_2$SO$_4$ to phenolphthalein end point and 2.5 ml more to methyl orange. Find the type and extent of alkalinity as CaCO$_3$.
Last-minute revision
- Hardness is expressed as CaCO$_3$ equivalent: mass $\times$ 50 / Eq. wt.
- 1 ppm = 1 mg/L = 0.07 °Cl = 0.1 °Fr.
- Temporary hardness: bicarbonates of Ca and Mg, removed by boiling.
- Permanent hardness: chlorides and sulphates of Ca and Mg, not removed by boiling.
- EDTA titration: pH 10 buffer, EBT indicator, wine red to blue.
- Alkalinity is due to OH$^-$, CO$_3^{2-}$, HCO$_3^-$; indicators phenolphthalein (P) and methyl orange (M).
- OH$^-$ and HCO$_3^-$ never coexist.
- $P > M/2$: OH$^-$ = $2P - M$, CO$_3^{2-}$ = $2(M-P)$.
- Ozone decomposes as $\text{O}_3 \rightarrow \text{O}_2 + [\text{O}]$.
- Paper numericals: 19 ppm; 720 / 1440 ppm; 175 + 50 ppm.
Memory hooks
- P and M: "P is Pink to clear, M is Yellow to red".
- P = 0 means only bicarbonate; P = M means only hydroxide.
- Temporary = two "b"s: bicarbonate, boiling.
- EBT: "Red to Blue is Ready" at end point.
Coverage checklist
- Sources: no past question.
- Impurities: Nov 2022 ozonation (7 marks).
- Hardness & its units: Jun 2022 units, Dec 2024 temporary vs permanent.
- Determination of hardness by EDTA method: Nov 2022 principle and estimation.
- Alkalinity & It’s determination: Jun 2022 method, Dec 2023 and Dec 2024 notes.
- related numerical problems: Nov 2022 / Jun 2025 hardness, Dec 2023 EDTA, Jun 2025 alkalinity.