Question

230 mL of 0.0666 mol/L Fe(NO3)3 is added to 320 mL of 0.010 mol/L KOH. Does a precipitate form?

Answer 1

1) $moles (Fe(NO_3)_3)= 0.0666 \frac{mol}{L} \times 0.230 L = 0.01532 mol$

$molarity (Fe(NO_3)_3)= \frac{0.01532 moles}{(0.230+0.320) L}=0.02785 \frac{mol}{L}$

2) $moles (KOH) = 0.010 \frac{mol}{L} \times 0.320 L = 0.0032 moles$

m$molarity (KOH) =\frac{0.0032 moles}{(0.230+0.320)L} = 0.005818 \frac{mol}{L}$

3) Chemical equation

$Fe(NO_3)_3 (aq) + 3KOH(aq) \rightarrow Fe(OH)_3(s) + 3KNO_3(aq)$

$Fe^{3+} (aq) + 3OH^-(aq) \rightarrow Fe(OH)_3(s)$

$Q_{sp} = [Fe^{3+}][OH^-]^3= [0.02785]\times[0.005818]^3=5.48\times10^{-9}$

$K_{sp} = 6\times 10^{-38}$

As $Q_{sp}>K_{sp}$ , a precipitate is formed.