---
title: "Alkalinity and Ca, Mg: TH 280, CH 75"
subject: "Water Supply Engineering"
chapter: 4
key: "semester_5/water_supply_engineering:hardness-s73"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/hardness-s73
site: BCT Engineering
---

# Alkalinity and Ca, Mg: TH 280, CH 75

Water Supply Engineering · Chapter 4: Quality of Water

- Asked in: 2073 Shrawan, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/hardness-s73
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-4

## Question

Determine total alkalinity and concentration of calcium and magnesium in the water sample if both calcium and magnesium ions were found equal. Total hardness is 280 mg/l and carbonate hardness is 75 mg/l.

## Short answer

Total alkalinity is 75 mg/l as CaCO3 because carbonate hardness is less than total hardness. With Ca and Mg equal, solve 280 = x(2.5 + 4.1), giving Ca = Mg = 42.36 mg/l.

## Answer

Total hardness TH $= 280$ mg/l, carbonate hardness CH $= 75$ mg/l, Ca$^{2+}$ = Mg$^{2+}$ $= x$ mg/l.

#### Total alkalinity
TH is greater than CH, so the alkalinity is fully used in carbonate hardness: total alkalinity = CH $= 75$ mg/l as CaCO$_3$. (Non-carbonate hardness $= 280-75 = 205$ mg/l.)

#### Concentration of Ca and Mg
$$
\begin{aligned}
\text{TH} &= x\left(\frac{50}{20.04}+\frac{50}{12.15}\right) = 6.610x = 280\\
x &= 42.36\ \text{mg/l}
\end{aligned}
$$

**Answer: Total alkalinity = 75 mg/l as CaCO$_3$; Ca$^{2+}$ = Mg$^{2+}$ = 42.36 mg/l.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Total alkalinity equals:
   A. CH = 75 mg/l as CaCO3
   B. TH = 280
   C. NCH = 205
   D. 355
   Answer: A. TH > CH so alkalinity = CH. (Key point: Alkalinity)

2. NCH = 280 - 75 =
   A. 355 mg/l
   B. 205 mg/l
   C. 75 mg/l
   D. 280 mg/l
   Answer: B. TH minus CH. (Key point: NCH)

3. With Ca = Mg = x, TH =
   A. 2.495x
   B. 4.115x
   C. 6.610x
   D. 7.752x
   Answer: C. 50/20.04 + 50/12.15. (Key point: Equation)

4. x = 280/6.610 =
   A. 56.0 mg/l
   B. 140 mg/l
   C. 11.3 mg/l
   D. 42.36 mg/l
   Answer: D. Each of Ca and Mg. (Key point: Concentration)

## Related in Water Supply Engineering

- [Ca and Mg concentration, NCH, alkalinity: TH 400 ppm](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ahardness-bk79) — Water Supply Engineering, chapter 4
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