{"id":98,"date":"2022-07-13T17:38:33","date_gmt":"2022-07-13T17:38:33","guid":{"rendered":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/chapter\/solution-exercise-6\/"},"modified":"2022-07-18T01:00:22","modified_gmt":"2022-07-18T01:00:22","slug":"solution-exercise-6","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/chapter\/solution-exercise-6\/","title":{"raw":"Solution Exercise 6","rendered":"Solution Exercise 6"},"content":{"raw":"
\r\n

For the Ethiopian Rift Valley sample, the molal concentration product, that is, the product of the molality of Ca2+<\/sup> times the molality of F-<\/sup> squared is as follows (note: calcium has 40 atomic mass units and fluorite has 19):<\/p>\r\n

[latex]\\displaystyle \\left ( a_{Ca^{2+}}a_{F^{-}}^{2} \\right )_{sample}=\\left ( 14.1\\frac{\\mathrm{mg}}{\\mathrm{L}}\\frac{1\\ \\mathrm{L}}{1000\\ \\mathrm{mg}}\\frac{1\\ \\mathrm{atom}}{40\\ \\mathrm{amu}} \\right )[\/latex] \u00d7 [latex]\\displaystyle \\left ( 2.62\\frac{\\mathrm{mg}}{\\mathrm{L}}\\frac{1\\ \\mathrm{L}}{1000\\ \\mathrm{mg}}\\frac{1\\ \\mathrm{atom}}{19\\ \\mathrm{amu}} \\right )^{2}=6.7\\times 10^{-12}[\/latex]<\/p>\r\n

Taking the logarithm results in log molal product\u00a0=\u00a0\u20131.17<\/p>\r\n

SI<\/em>\u00a0=\u00a0log\u00a0molal product\u00a0\u2013\u00a0log\u00a0K<\/em>sp<\/em><\/sub>\u00a0=\u00a0\u201311.17\u00a0\u2013\u00a0(\u201310.6)\u00a0=\u00a0\u20130.57<\/p>\r\n

This suggests undersaturated conditions. Using the PHREEQC code, SI<\/em> = \u22120.66 which is slightly more undersaturated. The difference is caused by the use of activity coefficients in the code calculation.<\/p>\r\n

Return to Exercise 6<\/a><\/p>\r\n

<\/p>\r\n\r\n<\/div>","rendered":"

\n

For the Ethiopian Rift Valley sample, the molal concentration product, that is, the product of the molality of Ca2+<\/sup> times the molality of F–<\/sup> squared is as follows (note: calcium has 40 atomic mass units and fluorite has 19):<\/p>\n

[latex]\\displaystyle \\left ( a_{Ca^{2+}}a_{F^{-}}^{2} \\right )_{sample}=\\left ( 14.1\\frac{\\mathrm{mg}}{\\mathrm{L}}\\frac{1\\ \\mathrm{L}}{1000\\ \\mathrm{mg}}\\frac{1\\ \\mathrm{atom}}{40\\ \\mathrm{amu}} \\right )[\/latex] \u00d7 [latex]\\displaystyle \\left ( 2.62\\frac{\\mathrm{mg}}{\\mathrm{L}}\\frac{1\\ \\mathrm{L}}{1000\\ \\mathrm{mg}}\\frac{1\\ \\mathrm{atom}}{19\\ \\mathrm{amu}} \\right )^{2}=6.7\\times 10^{-12}[\/latex]<\/p>\n

Taking the logarithm results in log molal product\u00a0=\u00a0\u20131.17<\/p>\n

SI<\/em>\u00a0=\u00a0log\u00a0molal product\u00a0\u2013\u00a0log\u00a0K<\/em>sp<\/em><\/sub>\u00a0=\u00a0\u201311.17\u00a0\u2013\u00a0(\u201310.6)\u00a0=\u00a0\u20130.57<\/p>\n

This suggests undersaturated conditions. Using the PHREEQC code, SI<\/em> = \u22120.66 which is slightly more undersaturated. The difference is caused by the use of activity coefficients in the code calculation.<\/p>\n

Return to Exercise 6<\/a><\/p>\n

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