The equation for the solubility equilibrium is
#"Fe(IO"_3)_3"(s)" ⇌ "Fe"^"3+""(aq)" + "3IO"_3^"-""(aq)"; K_text(sp) = 1.0 × 10^"-14"#
To determine whether a precipitate will form, we must calculate the ionic concentrations at the moment of mixing.
Step 1. Calculate the initial concentrations
The total volume after mixing is 30.00 mL.
Each solution is diluted from 15.00 mL to 30.00 mL, so the new concentrations are:
#["Fe"^"3+"] = 3.0 × 10^"-6"color(white)(l)"mol/L" × (15.00 color(red)(cancel(color(black)("mL"))))/(30.00 color(red)(cancel(color(black)("mL")))) = 1.50 × 10^"-6"color(white)(l)"mol/L"#
#["IO"_3^"-"] = 3.0 × 10^"-4"color(white)(l)"mol/L" × (15.00 color(red)(cancel(color(black)("mL"))))/(30.00 color(red)(cancel(color(black)("mL")))) = 1.50 × 10^"-4"color(white)(l)"mol/L"#
Step 2. Calculate #Q_text(sp)#
#color(white)(mmmmm)"Fe(IO"_3)_3"(s)" ⇌ "Fe"^"3+""(aq)" + color(white)(l)"3IO"_3^"-""(aq)"#
#"I/mol·L"^"-1": color(white)(mmmmmmm)1.50 ×10^"-6"color(white)(m)1.50 × 10^"-4"#
#Q_text(sp) = ["Fe"^"3+"]["IO"_3^"-"]^3 = (1.50 ×10^"-6")(1.50 × 10^"-4")^3 = 5.1 × 10^"-18"#
Step 3. Will a precipitate form?
#Q_text(sp) = 5.1 × 10^"-18"#
#K_text(sp) = 1.0 × 10^"-14"#
#"Q_text(sp) < K_text(sp)#
The concentrations of ions are not high enough to form a precipitate.
No precipitate will form.
