Ion concentrations obtained for a groundwater sample (having pH−8.10) are given below.

$\begin{matrix} Ion & C a^{2 +} & M g 2 + & N a^{+} & H C O_{3}^{-} & S O_{4}^{2 -} & C l^{-} Ion concentration (m g / L) & 100 & 6 & 15 & 250 & 45 & 39 Atomic weight & C a = 40 & M g = 24 & N a = 23 & H = 1 C = 12 Q = 16 & S = 32 O = 16 & C l = 35.5 \end{matrix}$ Carbonate hardness $(m g / L a s C a C O_{3})$ present in the above sample is

289

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205

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250

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275

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Solution

The correct option is B $205$
Carbonate hardness is equal to total harndess of alkalinity whichever is less.
Alkalintiy is caused by $C O_{3}^{2 -}, H C O_{3}^{-}, O H^{-}$
Hence, alkainity
$= [H C O_{3}^{-}] \times \frac{Equivalent weight of C a C O_{3}}{Equivalent weight of H C O_{3}^{-}}$
$= 250 \times \frac{50}{61} = 205 m g / L$
Hence carbonate hardness $= m i n {275, 205} m g / L$
$= 205 m g / L$ as $C a C O_{3}$

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Similar questions

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})

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

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

\begin{matrix} Ion & Concentration m g / l & Atomic Weight C a^{2 +} & 60 & 40 M g^{2 +} & 30 & 24.31 H C O_{3}^{-} & 400 & 61 \end{matrix}

C a C O_{3}

\begin{matrix} Ion & Concentration m g / l & Atomic Weight C a^{2 +} & 20.0 & 60 M g^{2 +} & 12.2 & 36.6 N a^{+} & 23.0 & 92 K^{+} & 39.1 & 78.2 S O_{4}^{2 -} & 48.0 & 72 H C O_{3}^{-} & 61.0 & 122 \end{matrix}

C a C O_{3}

C a C O_{3} = 50 m g / m e q

C a C O_{3}

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