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Byju's Answer

Standard XII

Chemistry

Valency and Oxidation State

Ion concentra...

Question

Ion concentrations obtained for a groundwater sample (having $p H = 8.10$ ) are given below.

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 $C a = 40$ $M g = 24$ $N a = 23$ $H = 1$
$C = 12$
$O = 16$ $S = 32$
$O = 16$ $C l = 35.5$

Total hardness $(m g / L a s C a C O_{3})$ present in the above water sample is

250

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275

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205

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308

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Solution

The correct option is B $275$
Total hardness in water is due to multivalent cations

Total hardness
$= [C a^{2 +}] \times \frac{Equivalent weight of C a C O_{3}}{Equivalent weight of C a^{2 +}} + [M g^{2 +}] \times \frac{Equivalent weight of C a C O_{3}}{Equivalent weight of M g^{2 +}}$

$= 100 \times \frac{50}{20} + 6 \times \frac{50}{12}$

$= 275 m g / L a s C a C O_{3}$

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Similar questions

Q. Ion concentrations obtained for a groundwater sample (having

p H - 8.10

) are given below.

\begin{matrix} 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 Q = 16 & S = 32 O = 16 & C l = 35.5 \end{matrix}

Total hardness

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

present in the above water sample is

Q. Ion concentration in water sample obtained from well is tabulated below

Ion	Ca++	Mg++	Na+	$H C O_{3}^{-}$	$S O_{4}^{- -}$
Concentration $(\frac{m g}{l})$	84	10	16	220	42

Total hardness (in

\frac{m g}{l} a s C a C O_{3})

of the water sample is____

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

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})

Q. 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}

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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	$C a = 40$	$M g = 24$	$N a = 23$	$H = 1$ $C = 12$ $O = 16$	$S = 32$ $O = 16$	$C l = 35.5$

The correct option is B 275Total hardness in water is due to multivalent cations Total hardness =[Ca2+]×Equivalent weight of CaCO3Equivalent weight of Ca2++[Mg2+]×Equivalent weight of CaCO3Equivalent weight of Mg2+ =100×5020+6×5012 =275mg/LasCaCO3