---
title: "Hardness with Ca, Mg, Sr equal, TH 150"
subject: "Water Supply Engineering"
chapter: 4
key: "semester_5/water_supply_engineering:hardness-as70"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/hardness-as70
site: BCT Engineering
---

# Hardness with Ca, Mg, Sr equal, TH 150

Water Supply Engineering · Chapter 4: Quality of Water

- Asked in: 2070 Asar, 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-as70
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-4

## Question

The total hardness value obtained from the analysis of a water sample is 150 mg/l. If all three (Ca, Mg and Sr) cations concentration causing hardness are numerically same and carbonate hardness is 77 mg/l, calculate the following: (i) the value of non carbonate hardness, (ii) the concentration of principal cations and (iii) the value of total alkalinity in mg/l.

## Short answer

Non-carbonate hardness is 150 minus 77 = 73 mg/l as CaCO3. With Ca = Mg = Sr = x, 150 = (2.5 + 4.1 + 1.14)x, so each cation is 19.38 mg/l. Total alkalinity equals carbonate hardness, 77 mg/l as CaCO3.

## Answer

Hardness is expressed as CaCO$_3$ equivalent. Each cation of concentration $C$ (mg/l) contributes $C \times \dfrac{50}{\text{eq. wt}}$ to hardness, which gives the standard factors: Ca 2.5, Mg 4.1, Sr 1.14.

$$
TH = 2.5\,Ca + 4.1\,Mg + 1.14\,Sr
$$

#### (i) Non-carbonate hardness
$$
NCH = TH - CH = 150 - 77 = 73\ \text{mg/l as CaCO}_3
$$

#### (ii) Concentration of principal cations
Let $Ca = Mg = Sr = x$.

$$
\begin{aligned} 150 &= (2.5 + 4.1 + 1.14)\,x = 7.74\,x \\ x &= \frac{150}{7.74} = 19.38\ \text{mg/l} \end{aligned}
$$

So Ca = Mg = Sr = 19.38 mg/l each.

#### (iii) Total alkalinity
Carbonate hardness is the part of hardness balanced by alkalinity. When alkalinity is less than total hardness, carbonate hardness equals alkalinity (the alkalinity is due to bicarbonate). Since CH = 77 < TH = 150:

$$
\text{Total alkalinity} = CH = 77\ \text{mg/l as CaCO}_3
$$

**Answer: NCH = 73 mg/l; Ca = Mg = Sr = 19.38 mg/l; total alkalinity = 77 mg/l (as CaCO$_3$).**

## Check yourself

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

1. How is NCH found from TH and CH?
   A. NCH = TH / CH
   B. NCH = CH - TH
   C. NCH = TH + CH = 227 mg/l
   D. NCH = TH - CH = 73 mg/l
   Answer: D. Total hardness is carbonate plus non-carbonate, so NCH = 150 - 77 = 73 mg/l. (Key point: Non-carbonate hardness)

2. With Ca = Mg = Sr = x, which equation gives x?
   A. 150 = (2.5 x 4.1 x 1.14)x
   B. 150 = 2.5x, x = 60 mg/l
   C. 150 = 3x, x = 50 mg/l
   D. 150 = (2.5 + 4.1 + 1.14)x, x = 19.38 mg/l
   Answer: D. The hardness factors add: 7.74x = 150, so x = 19.38 mg/l each. (Key point: Cation concentration)

3. Why is total alkalinity 77 mg/l?
   A. Alkalinity always equals TH
   B. CH (77) is less than TH, so CH equals the alkalinity
   C. Alkalinity is 150 - 73 x 2
   D. Alkalinity equals NCH
   Answer: B. When alkalinity is less than total hardness, carbonate hardness equals the alkalinity. (Key point: Total alkalinity)

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- [Non-carbonate hardness: TH 150, CH 60](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ahardness-c72) — Water Supply Engineering, chapter 4, asked 2×
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