Question #fa00c

A solution with pH 8.5 has a pOH of 14-8.5 = 5.5, therefore $\left[O {H}^{-}\right]$= $3.16 \cdot {10}^{-} 6$. The dilution of 12.5 ml to 25 ml brings this concentration to $\left[O {H}^{-}\right]$= $1.58 \cdot {10}^{-} 6$.
$\left[A {l}^{3 +}\right] \cdot {\left[O {H}^{-}\right]}^{3}$ = $\frac{3.5 \cdot {10}^{-} 8}{2} \cdot {\left(1.58 \cdot {10}^{-} 6\right)}^{3}$=$1.75 \cdot {10}^{-} 8 \cdot 4.74 \cdot {10}^{-} 18$= $8.29 \cdot {10}^{-} 26$
This value exceeds the ksp of $3.1 \cdot {10}^{-} 34$, and a precipitation will occur.