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 $m g / L$ as $C a C O_{3}$

5000

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250

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1750

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1500

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Solution

The correct option is C $1750$
Alkalinity is due to $(C O_{3}^{-}, H C O_{3}^{-} O H^{-})$ [major source]
Alkalinity = Milliequivalent of $C O_{3}^{2 +} +$ Milliquivalent of $H C O_{3}^{-}$
$= 05 + 30$
$= 35$ milliequivalent/L
$= (35 \times e q . w t . o f C a C O_{3} / L) m g / L$
$= 35 \times 50 m g / L$
$= 1750 m g / L a s C a C O_{3}$

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Similar questions

\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}

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

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 $C a C O_{3}$

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