Following chemical species were reported for water sample from a well

Species Concentration
(milliequivalent/L)

Chloride $(C l^{-})$ 15

Sulphate $(S O_{4}^{2 -})$ 15

Carbonate $(C O_{3}^{2 -})$ 05

Bicarbonate $(H C O_{3}^{-})$ 30

Calcium $(C a^{2 +})$ 12

Magnesium $(M g^{2 +})$ 18

$p H$ 8.5

Total hardness in mg/L as $C a C O_{3}$ is

5000

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2000

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1500

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3000

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Solution

The correct option is C $1500$
Total hardness is due to multivalent cations $(C a^{2 +}, M g^{2 +})$

Total hardess
$=$ milliequivalent of $C a^{2 +}$ $+$ milliequivalent of $M g^{2 +}$
$= 12 + 18$

=30 milliequivalent/L
=30 milliequivalent $C a C O_{3}$
$= 30 \times e q u i v a l e n t w e i g h t o f C a C O_{3} m g / L$
$= 30 \times 50 m g / L$
$= 1500 m g / L a s C a C O_{3}$

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Similar questions

Following chemical species were reported for water sample from a well

Species	Concentration (milliequivalent/L)
Chloride $(C l^{-})$	15
Sulphate $(S O_{4}^{2 -})$	15
Carbonate $(C O_{3}^{2 -})$	05
Bicarbonate $(H C O_{3}^{-})$	30
Calcium $(C a^{2 +})$	12
Magnesium $(M g^{2 +})$	18
$p H$	8.5

Alkalinity present in the water in mg/L as $C a C O_{3}$

Q. Following chemical species were reported for water sample from a well

\begin{matrix} Species & Concentration (milliequivalent/L) Chloride (C l^{-}) & 15 Sulphate (S O_{4}^{2 -}) & 15 Carbonate (C O_{3}^{2 -}) & 05 Bicarbonate (H C O_{3}^{-}) & 13 Calcium (C a^{2 +}) & 12 Magnesium (M g^{2 +}) & 18 p H & 8.5 \end{matrix}

Total hardness in

m g / L a s C a C O_{3}

Q. Following chemical species were reported for water sample from a well

\begin{matrix} Species & Concentration (milliequivalent/L) Chloride (C l^{-}) & 15 Sulphate (S O_{4}^{2 -}) & 15 Carbonate (C O_{3}^{2 -}) & 05 Bicarbonate (H C O_{3}^{-}) & 13 Calcium (C a^{2 +}) & 12 Magnesium (M g^{2 +}) & 18 p H & 8.5 \end{matrix}

Alkalinity present in the water in

m g / L

C a C O_{3}

Q. A water sample or testing gave the following data
From the above data , non carbonate
hardness in the water sample is use
atomic weights as

[N a^{+}] = 38 m g / l, ∣ ∣ C a^{2_{+}} ∣ ∣ = 10 m g / l

[H C O_{3}^{-}] = 183 m g / l, [M g^{2 +}] = 12 m g / l

[C O_{3}^{2 -}] = 30 m g / l, [A l^{3 +}] = 27 m g / l

[S O_{4}^{2 -}] = 45 m g / l

Ca = 40, O = 16 , Mg = 24 , Na = 23 , Al = 27, C = 12, H = 1, S = 32

(i n m g / l a s C a C O_{3})

A water contains the following dissolved ions.
$[N a^{+}] = 56 m g / L; [C a^{2 +}] = 40 m g / L;$
$[M g^{2 +}] = 30 m g / L; [A l^{3 +}] = 3 m g / L;$
$[H C O_{3}^{-}] = 190 m g / L; [C l^{-}] = 165 m g / L;$
Water $p H$ is $7$ .
Atomic weights:
$C a = 40, M g = 24, A l = 27, H = 1, C = 12, O = 16, N a = 23, C l = 35.5$

The non-carbonate hardness of the sample in $m g / L$ as $C a C O_{3}$ is

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Following chemical species were reported for water sample from a well Species Concentration (milliequivalent/L) Chloride (Cl−) 15 Sulphate (SO2−4) 15 Carbonate (CO2−3) 05 Bicarbonate (HCO−3) 30 Calcium (Ca2+) 12 Magnesium (Mg2+) 18 pH 8.5 Total hardness in mg/L as CaCO3 is

The correct option is C 1500Total hardness is due to multivalent cations (Ca2+,Mg2+) Total hardess = milliequivalent of Ca2+ + milliequivalent of Mg2+ =12+18 =30 milliequivalent/L =30 milliequivalent CaCO3 =30× equivalent weight of CaCO3 mg/L =30×50mg/L =1500mg/LasCaCO3