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Chapter 21: Problem 67

A particular water sample contains 56.9 ppm \(\mathrm{SO}_{4}^{2-}\) and \(176 \mathrm{ppm} \mathrm{HCO}_{3}^{-},\) with \(\mathrm{Ca}^{2+}\) as the only cation. (a) How many parts per million of \(\mathrm{Ca}^{2+}\) does the water contain? (b) How many grams of \(\mathrm{CaO}\) are consumed in removing \(\mathrm{HCO}_{3}^{-}\), from \(602 \mathrm{kg}\) of the water? (c) Show that the \(\mathrm{Ca}^{2+}\) remaining in the water after the treatment described in part (b) can be removed by adding \(\mathrm{Na}_{2} \mathrm{CO}_{3}.\) (d) How many grams of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) are required for the precipitation referred to in part (c)?

Short Answer

Expert verified
The water contains 23.7 ppm of Ca2+, 660 g of CaO are consumed in removing HCO3- from 602 kg of the water, the remaining Ca2+ in the water can be removed by adding Na2CO3, and 1418 g of Na2CO3 are required for the precipitation.

Step by step solution

01

Calculate Ca2+ in ppm

The molar masses of SO4^2- and Ca2+ are 96 g/mol and 40 g/mol respectively. The ppm (parts per million) is a unit of concentration. Assuming that SO4^2- and Ca2+ are in a 1:1 ratio (because CaSO4 is formed), the concentration of Ca2+ would be \(Ca^{2+} = \frac{56.9 \times 40}{96} = 23.7 ppm\).
02

Calculate amount of CaO for removing HCO3-

The formation of CaCO3 from CaO and HCO3- is in a 1:1 ratio. Therefore, for each mole of HCO3-, one mole of CaO is needed. Given that the molar mass of CaO is 56 g/mol and the ppm of HCO3- is 176 (which means 176 mg/kg), we have: \[CaO = \frac{176 \times 602 \times 56}{96} = 660 g\] of CaO consumed in removing HCO3- from 602 kg of water.
03

Prove that remaining Ca2+ can be removed by Na2CO3

The reaction between Ca2+ and Na2CO3 leads to formation of CaCO3 and 2Na+. This reaction proves that the remaining Ca2+ in the water can be precipitated out by adding Na2CO3.
04

Calculate required amount of Na2CO3

Again using stoichiometry, for each mole of Ca2+ one mole of Na2CO3 is needed. Given that the molar mass of Na2CO3 is 106 g/mol and the ppm of Ca2+ left in the water after removal of HCO3- is 23.7 (which means 23.7 mg/kg), we have: \[Na2CO3 = \frac{23.7 \times 602 \times 106}{40} = 1418 g\] of Na2CO3 required to precipitate the remaining Ca2+.

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