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

2000

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1500

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3000

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5000

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $1500$
Total hardness is due to multivalent cations $(C a^{2 +}, M g^{2 +})$
Total harndess $=$ milliequivalent of $C a^{2 +}$ $+$ milliequivalent of $M g^{2 +}$
$= 12 + 18$
$= 30$ milliequivalent/L
$= 30$ milliequivalent $C a C O_{3}$
$= 30 \times$ equivalent weight of $C a C O_{3}$ $m g / L$
$= 30 \times 50 m g / L$
$= 1500 m g / L a s C a C O_{3}$

Suggest Corrections

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

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

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 $C a C O_{3}$

[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

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

p H = 8.10

Ion	$C a^{2 +}$	$M g^{2 +}$	$N a^{+}$	$H C O_{3}^{-}$	$S O_{4}^{2 -}$	$C l^{-}$
Ion concentration $(m g / L)$	$100$	$6$	$15$	$250$	$45$	$39$
Atomic weight	$C a = 40$	$M g = 24$	$N a = 23$	$H = 1$ $C = 12$ $O = 16$	$S = 32$ $O = 16$	$C l = 35.5$

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

Join BYJU'S Learning Program