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the ph at which a 0.01 M Al+3 solution is 99.99 % precipitated is (Al(OH)3Ksp : 1 .99*10^-9)Answer is 10
Al(OH)3 === Al3+ + 3OH- s 0 0 0 s 3s s.(3s)3 = 10-18 27s4 = 10-18 s =~ 10-4/7Now, OH- = 3sOH- = 3x10-4/7pOH = -log 3x10-4/7pOH = 3.8pH = 14-pOHpH = 10.3
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