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Standard XII

Chemistry

Basic Buffer Action

1000 ml of ta...

Question

1000 ml of tap water containing $C a (H C O_{3})_{2}$ was titrated with $\frac{N}{50}$ HCl with methyl orange as indicator. If 300 ml of HCl were required, calculate temporary hardness as parts of $C a C O_{3} p e r 10^{6}$ parts of water

150 ppm

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300 ppm

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450 ppm

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600 ppm

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Solution

The correct option is B 300 ppm
$m . e q o f C a C O_{3} = m . e q o f C a (H C O_{3})_{2} = m . e q H C l = 300 \times \frac{1}{50} = 6$
$∴ w t o f C a C O_{3} = 6 \times \frac{100}{2} m g . L^{- 1} = 300 p p m$

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Similar questions

Q. 100 ml of tap water containing

C a (H C O_{3})_{2}

was titrated with N/50

H C l

with methyl orange as indicator. If 30ml of

H C l

were required, calculate the temporary hardness as part of

C a C O_{3}

per

10^{6}

parts of water.

Q. 20 liter of tap water that has

C a (H C O_{3})_{2}

was treated with 90 ml of M/50 HCl. The resulting solution requires an extra 6 ml of 0.1 M NaOH to make the solution neutral. What is the total hardness as parts of

C a C O_{3}

in ppm? (Take the molar mass of

C a (H C O_{3})_{2}

= 162 g/mol)

Q. Hardness of water is measured in terms of ppm (parts per million) of

C a C O_{3}

. It is the amount (in g) of

C a C O_{3}

present in

10^{6} g H_{2} O

. In a sample of water,

10

L required

0.56

g of

C a O

to the remove temporary hardness of

H C O_{3}^{-}

C a (H C O_{3})_{2} + C a O ⟶ 2 C a C O_{3} + H_{2} O

Temporary hardness is :

10

L of hard water with temporary hardness of

[C a (H C O_{3})_{2}]

requires

0.56

g of lime. The reaction follows as:

C a (H C O_{3})_{2} + C a O ⟶ 2 C a C O_{3} + H_{2} O

.
Temporary hardness in terms of ppm of

C a C O_{3}

is:

Q. A 20 ml solution of 0.1 N

N a_{2} C O_{3}

is titrated with 0.1N HCl with phenolphthalein as the indicator. Another 20 ml solution of 0.1 N

N a_{2} C O_{3}

is titrated with 0.1 N HCl with methyl orange as the indicator. The volume of HCl required for both the titrations resepectively is:

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1000 ml of tap water containing Ca(HCO3)2 was titrated with N50 HCl with methyl orange as indicator. If 300 ml of HCl were required, calculate temporary hardness as parts of CaCO3 per 106 parts of water

The correct option is B 300 ppmm.eq of CaCO3=m.eq of Ca(HCO3)2=m.eqHCl=300×150=6 ∴wt of CaCO3=6×1002mg.L−1=300 ppm

1000 ml of tap water containing $C a (H C O_{3})_{2}$ was titrated with $\frac{N}{50}$ HCl with methyl orange as indicator. If 300 ml of HCl were required, calculate temporary hardness as parts of $C a C O_{3} p e r 10^{6}$ parts of water

The correct option is B 300 ppm
$m . e q o f C a C O_{3} = m . e q o f C a (H C O_{3})_{2} = m . e q H C l = 300 \times \frac{1}{50} = 6$
$∴ w t o f C a C O_{3} = 6 \times \frac{100}{2} m g . L^{- 1} = 300 p p m$