Answer to Question #223324 in General Chemistry for SAmanta Ariel

Question

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 ?

Answer 1

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

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, $p = \log(x)$

$3x = 0.2(\log x²) × 40(\log x³)/1000\\ 3x = 0.2(2\log x) × 40(3\log x)/1000\\ 3x = 0.4 \log x × 0.12\log x\\ 3x =0.048\log x\\ \log x = 62.5\ x\\ x \approx -0.045x+ 0.039\ i$