Answer to Question #223324 in General Chemistry for SAmanta Ariel

Question #223324

What is the mass of Al2(SO4)3 needed to prepare 40 (p3)ml of (0.2p2)M of sulphate ions in aluminum sulphate solution? where p= log(moles of Al2(SO4)3 ?


1
Expert's answer
2021-11-12T14:56:02-0500

"Al_2(SO_4)_3 \\leftrightarrow 2Al^{3+} + 3SO_4^{2+}"

1 mole of Aluminium Sulphate solution produces 3 moles of Sulphate ions


x moles of Aluminium Sulphate solution produces 3x moles of Sulphate ions

3x = molarity × volume

3x = 0.2p² × 40p³/1000


Since, "p = \\log(x)"

"3x = 0.2(\\log x\u00b2) \u00d7 40(\\log x\u00b3)\/1000\\\\\n3x = 0.2(2\\log x) \u00d7 40(3\\log x)\/1000\\\\\n3x = 0.4 \\log x \u00d7 0.12\\log x\\\\\n3x =0.048\\log x\\\\\n\\log x = 62.5\\ x\\\\\nx \\approx -0.045x+ 0.039\\ i"

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