Permanent hardness is due to Cl- and SO2-4 of Mg2- and Ca2- and is removed by adding Na2CO3- CaSO4-Na2CO3- CaCO3-Na2SO4 CaCl2-Na2CO3- CaCO3-2NaClWhich of the following statements is-are correct-

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Byju's Answer

Standard XII

Chemistry

Hardness of Water and Treatment

Permanent har...

Question

Permanent hardness is due to $C l^{⊝}$ and $S O_{4}^{2 -}$ of $M g^{2 +}$ and $C a^{2 +}$ and is removed by adding $N a_{2} C O_{3}$ .

$C a S O_{4} + N a_{2} C O_{3} \to C a C O_{3} + N a_{2} S O_{4}$
$C a C l_{2} + N a_{2} C O_{3} \to C a C O_{3} + 2 N a C l$

Which of the following statements is/are correct?

If hardness is

100

ppm

C a C O_{3}

, the amount of

N a_{2} C O_{3}

required to soften

10

L of hard water is

1.06

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If hardness is

100

ppm

C a C O_{3}

, the amount of

N a_{2} C O_{3}

required to soften

10

L of hard water is

10.6

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If hardness is

420

ppm

M g C O_{3}

, the amount of

N a_{2} C O_{3}

required to soften

10

L of hard water is

53.0

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If hardness is

420

ppm

M g C O_{3}

, the amount of

N a_{2} C O_{3}

required to soften

10

L of hard water is

5.3

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Solution

The correct options are
A If hardness is $100$ ppm $C a C O_{3}$ , the amount of $N a_{2} C O_{3}$ required to soften $10$ L of hard water is $1.06$ g.
D If hardness is $420$ ppm $M g C O_{3}$ , the amount of $N a_{2} C O_{3}$ required to soften $10$ L of hard water is $5.3$ g.
Molecular weight of $C a C O_{3}$ $= 40 + 12 + 3 \times 16 = 100$ g/mol
Molecular weight of $N a_{2} C O_{3} = 46 + 60 = 106$ g/mol

i. $C a S O_{4} x m o l + N a_{2} C O_{3} x m o l \to C a C O_{3} x m o l + N a_{2} S O_{4}$

ii. $C a C l_{2} y m o l + N a_{2} C O_{3} y m o l \to C a C O_{3} y m o l + 2 N a C l$

iii. $100$ ppm $C a C O_{3} \equiv 100$ g $C a C O_{3}$ in $10^{6}$ mL $= \frac{100}{100}$ mol in $10^{6}$ mL

Moles of $N a_{2} C O_{3}$ required $=$ Moles of $C a C O_{3}$
$\equiv \frac{100}{100}$ mol in $10^{6}$ mL
$\equiv 1$ mol in $10^{6}$ mL
$\equiv \frac{1}{10^{6}} \times 10 \times 10^{3}$ mL per $10$ L $= 1 \times 10^{- 2}$ mol in $10$ L

Therefore, moles of $N a_{2} C O_{3}$ required
$= 1 \times 10^{- 2}$ mol in $10$ L
$= 1 \times 10^{- 2} \times 106$ g per $10$ L
$= 1.06$ g

Similarly, for $100$ ppm $M g C O_{3}$ (Molecular weight of $M g C O_{3} = 24 + 60 = 84$ g/mol),

iv. $M g S O_{4} x m o l + N a_{2} C O_{3} x m o l \to M g C O_{3} x m o l + N a_{2} S O_{4}$

v. $M g C l_{2} y m o l + N a_{2} C O_{3} y m o l \to M g C O_{3} y m o l + 2 N a C l$

vi. $420$ ppm $M g C O_{3} \equiv 420$ g $M g C O_{3}$ in $10^{6}$ mL
$\equiv \frac{420}{84} = 5$ mol in $10^{6}$ mL

Moles of $N a_{2} C O_{3}$ required $\equiv$ moles of $M g C O_{3}$
$\equiv 5$ mol in $10^{6}$ mL
$\equiv \frac{5}{10^{6}} \times 10 \times 10^{3}$ mL per $10$ L
$\equiv 5 \times 10^{- 2}$ mol in $10$ L

Moles of $N a_{2} C O_{3}$ required $\equiv 5 \times 10^{- 2}$ mol per $10$ L
$\equiv 5 \times 10^{- 2} \times$ per $10$ L
$= 5.3$ g per $10$ L

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Similar questions

Q. Permanent hardness is due to

S O_{4}^{2 -}

and

C l^{-}

C a^{2 +}

and

M g^{2 +}

and is removed by the addition of

N a_{2} C O_{3}

C a S O_{4} + N a_{2} C O_{3} ⟶ C a C O_{3} + N a_{2} S O_{4}

C a C l_{2} + N a_{2} C O_{3} ⟶ C a C O_{3} + 2 N a C l

If hardness is

100

ppm

C a C O_{3}

, amount of

N a_{2} C O_{3}

required to soften

10

L of hard water is :

Q. Presence of

C a S O_{4}

causes permanent hardness in water. It can be removed by precipitating it as

C a C O_{3}

by adding

N a_{2} C O_{3}

. If the amount of

N a_{2} C O_{3}

required to soften 100 L of hard water, which is 200 ppm

C a C O_{3}

, is

x

g, then value of

x

is:

N a_{2} C O_{3}

is widely used in softening of hard water. If 1 L of hard water required

0.0106 g

N a_{2} C O_{3}

, The hardness in ppm (parts per million i.e.,

10^{6}

ml) of

C a C O_{3}

is:

10

L of hard water required

0.56

g of lime

(C a O)

for removing hardness. Hence, temporary hardness in ppm of

C a C O_{3}

is :

Q. Washing soda

(N a_{2} C O_{3} .10 H_{2} O)

is widely used in softening of hard water. If

1

L of hard water requires

0.0286

g of washing soda, the hardness of

C a C O_{3}

in ppm is :

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Permanent hardness is due to Cl⊝ and SO2−4 of Mg2+ and Ca2+ and is removed by adding Na2CO3.CaSO4+Na2CO3→CaCO3+Na2SO4CaCl2+Na2CO3→CaCO3+2NaClWhich of the following statements is/are correct?