Answer to Question #141787 in General Chemistry for Shivam Sinha

Question #141787
Calculate in detail total temporary hardness, total permanent hardness, in terms of calcium
carbonate equivalents in a water sample containing Magnesium bicarbonate (15.5 mg), Calcium
Sulphate (8.5 mg), Magnesium Chloride (9.1 mg), calcium sulphate (13 mg)
1
Expert's answer
2020-11-02T09:03:05-0500

"\\begin{aligned}\nMg(HCO_3)_2 =& \\ 15.5mg\\\\\nCaSO_4 =&\\ 8.5mg\\\\\nMgCl_2 =&\\ 9.1g\\\\\nCa(HCO_3)_2 =&\\ 13mg\n\\end{aligned}"


Temporary hardness is present in bicarbonates only;

"\\begin{aligned}\n\\textsf{mole of }Mg(HCO_3)_2 &= \\dfrac{15.5\u00d710^{-3}}{146g\/mol} \\\\\n\\\\\n&= 1.06\u00d7 10^{-4} moles\\\\\n\\\\\n\\textsf{mole of }Ca(HCO_3)_2 &= \\dfrac{13\u00d710^{-3}}{162g\/mol} \\\\\n\\\\\n&= 8.02\u00d7 10^{-5} moles\n\\end{aligned}"

"\\begin{aligned}\n\\\\\n\\textsf{Total moles of }Ca^+\\textsf{and }Mg ^+ &= (1.06 \u00d7 10^{-4})+(8.02\u00d710^{-5})\\\\\n&=1.86\u00d710^{-4} moles\n\\end{aligned}"

"\\begin{aligned}\n\\\\\n\\therefore \\textsf{Total moles of }CaCO_3 = 1.86\u00d710^{-4} moles\n\\end{aligned}"

"\\begin{aligned}\n\\textsf{Total mass of } CaCO_3 =1.86\u00d71&0^{-4} moles \u00d7 100g\/mol\\\\\n=0.0186g&\\\\\n\\\\\n\\therefore \\textsf{ppm (temporary hardness)} &= \\dfrac{0.0186g}{1000} \u00d7 10^6\\\\\n\\\\\n&= 18.6ppm\n\\end{aligned}"



Permanent hardness is present in Chlorides and Sulphates;

"\\begin{aligned}\n\\\\\n\\textsf{mole of }MgCl_2 &= \\dfrac{9.1\u00d710^{-3}g}{95g\/mol} \\\\\n\\\\\n&= 9.58\u00d7 10^{-5} moles\\\\\n\\\\\n\\textsf{mole of }CaSO_4 &= \\dfrac{8.5\u00d710^{-3}g}{136g\/mol} \\\\\n\\\\\n&= 6.25\u00d7 10^{-5} moles\n\\end{aligned}"

"\\begin{aligned}\n\\\\\n\\textsf{Total moles of }Ca^+\\textsf{and }Mg ^+ &= (9.58 \u00d7 10^{-5})+(6.25\u00d710^{-5})\\\\\n&=1.58\u00d710^{-4} moles\n\\end{aligned}"

"\\begin{aligned}\n\\\\\n\\therefore \\textsf{Total moles of }CaCO_3 = 1.58\u00d710^{-4} moles\n\\end{aligned}"

"\\begin{aligned}\n\\textsf{Total mass of } CaCO_3 =1.58\u00d71&0^{-4} moles \u00d7 100g\/mol\\\\\n=0.0158g&\\\\\n\\\\\n\\therefore \\textsf{ppm (temporary hardness)} &= \\dfrac{0.0158g}{1000} \u00d7 10^6\\\\\n\\\\\n&= 15.8ppm\n\\end{aligned}"



"\\therefore" The total temporary hardness and the total permanent hardness in terms of Calcium Carbonate equivalents is 18.6ppm and 15.8ppm respectively.


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