A water containing 8.1 mg/L of Ca(HCO\( _3 \))\( _2 \) has a hardness of
Answer: C
The equivalent is 8.1 times 100 over 162, that is 5 parts per million as calcium carbonate. The salt's molar mass of 162 makes its factor a little under two thirds, and being a bicarbonate it contributes temporary hardness that boiling would remove.
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A water containing 12 mg/L of MgSO\( _4 \) has a hardness of
Answer: D
Twelve times 100 over 120 gives 10 parts per million as calcium carbonate, the factor being the ratio of the two molar masses. Magnesium sulphate contributes permanent hardness, so boiling will not remove it and soda ash is needed.
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A hardness of 100 ppm expressed in degrees Clark is about
Answer: A
Dividing by 14.3 gives 6.99, that is about 7 degrees Clark, since each such degree is a grain of calcium carbonate in an imperial gallon. The same water measures 10 degrees French, the French unit being a part in a hundred thousand.
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A hardness of 250 ppm expressed in degrees French is
Answer: B
Each French degree is 10 parts per million, so 250 divided by 10 gives 25. In Clark degrees the same water measures about 17.5, and the conversion between the two is simply the ratio 14.3 to 10. A water of this hardness is decidedly hard and needs softening before boiler use.
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50 mL of a water sample required 12 mL of EDTA, 1 mL of which is equivalent to 1 mg of CaCO\( _3 \). The total hardness is
Answer: C
Twelve millilitres of EDTA correspond to twelve milligrams of calcium carbonate equivalent in the 50 millilitre aliquot, which is 240 milligrams in a litre. Scaling the titration volume up to a litre is the step most often got wrong in this calculation.
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A water contains Mg(HCO\( _3 \))\( _2 \) 14.6 mg/L and CaSO\( _4 \) 13.6 mg/L. Its total hardness is
Answer: D
The first gives 14.6 times 100 over 146, that is 10, and the second 13.6 times 100 over 136, also 10, so the total is 20 parts per million. The bicarbonate contributes the temporary part and the sulphate the permanent part, in equal measure here.
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A water of temporary hardness 200 ppm and permanent hardness 100 ppm, all as CaCO\( _3 \), needs lime for the temporary part amounting to
Answer: A
Correct answer: 148 mg/L
Removing temporary hardness needs one mole of calcium hydroxide per mole of hardness, so the requirement is the hardness expressed as calcium carbonate scaled by the ratio of the molar masses: \[ 200 \times \frac{74}{100} = 148\ \mathrm{mg/L}. \] The 74 is the molar mass of \( \mathrm{Ca(OH)_2} \) and the 100 that of \( \mathrm{CaCO_3} \). The permanent hardness needs soda ash, not lime, so it plays no part in this calculation.
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For that same water the soda ash needed for the permanent hardness is
Answer: B
The soda requirement is 106 over 100 of the permanent hardness, so 100 times 1.06 gives 106 milligrams per litre. Sodium carbonate's molar mass of 106 against calcium carbonate's 100 supplies the factor, and only the permanent part needs it.
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Softening 50 000 litres of that water requires soda ash amounting to
Answer: C
One hundred and six milligrams per litre on fifty thousand litres is 5.3 million milligrams, that is 5.3 kilograms of soda ash. The lime requirement over the same volume, at 148 milligrams per litre, would be 7.4 kilograms.
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A zeolite bed of capacity 500 mg CaCO\( _3 \) equivalent per litre of bed, holding 20 litres of zeolite, will soften water of 250 ppm hardness to the extent of
Answer: D
The bed holds 10 000 milligrams of equivalent, and each litre of this water brings 250 milligrams, so forty litres exhaust it. The bed must then be taken off line and regenerated with a ten per cent brine solution before it will soften anything further.
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