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Standard XII

Chemistry

Hardness of Water and Treatment

100 ml of tap...

Question

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.

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
Milli equivalent of $C a (H C O_{3})_{2}$ = miili equivalent of $C a C O_{3}$ = mili equivalent of $H C l$
$N_{1} V_{1} = N_{2} V_{2}$
$1000 \times w$ /( $100$ / $2$ )= $30$ / $50$
$w$ = $0.03$ g
For $100$ ml the amount of $C a C O_{3} = 0.03$ g
For $10^{6}$ = $0.03 \times 10^{6}$ / $100 = 300$ ppm

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Similar questions

Q. 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

500 m L

of a solution of

C a C O_{3}

and

N a O H

was first titrated with

\frac{N}{10} H C l

using phenolphthalein indicator.

18.5 m L

H C l

was required for the end point. After that methyl orange indicator was used and

2.5 m L

of same

H C l

was required for the next end point. Find out the mass of

C a C O_{3}

in the mixture.

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:

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:

25 m L

of a mixture of

N a O H

and

N a_{2} C O_{3}

when titrated with

N / 10 H C l

using phenolphthalein indicator required

25 m L H C l

. The same volume of mixture when titrated with

N / 10 H C l

using methyl orange indicator required

30 m L

H C l

. Calculate the amount of

N a_{2} C O_{3}

and

N a O H

in one litre of this mixture.

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100 ml of tap water containing Ca(HCO3)2 was titrated with N/50 HCl with methyl orange as indicator. If 30ml of HCl were required, calculate the temporary hardness as part of CaCO3 per 106 parts of water.

The correct option is B 300 ppmMilli equivalent of Ca(HCO3)2= miili equivalent of CaCO3= mili equivalent of HClN1V1=N2V21000×w/(100/2)= 30/50w= 0.03 gFor 100ml the amount of CaCO3=0.03 gFor 106 = 0.03×106/100=300 ppm

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.