A water sample analysis data is given below:

Ion Concentration $m g / l$ Atomic Weight

$C a^{2 +}$ 60 40

$M g^{2 +}$ 30 24.31

$H C O_{3}^{-}$ 400 61

The carbonate hardness (expressed as mg/L, of $C a C O_{3}$ , up to one decimal place) for the water sample is
273.4

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Solution

The correct option is A 273.4
Total hardness
$= [C a^{+ +} (m g / l) \times \frac{50}{(40 / 2)} + M g^{+ +} (m g / l) \times \frac{50}{(24.31 / 2)}]$
$= [60 \times \frac{50}{20} + 30 \times \frac{50}{12.155}]$
$= 273.4 m g / l a s C a C O_{3}$

Alkanity
$= [H C O_{3}^{-} (m g / l) \times \frac{50}{61}]$
$= 400 \times \frac{50}{61}$
$= 327.87 m g / l a s C a C O_{3}$

Carbonate hardness $=$ Minimum (total hardness, alkanity)

$= 273.4 m g / l a s C a C O_{3}$ .

Suggest Corrections

Similar questions

\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}

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})

C a C O_{3}

C a C O_{3}

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