Question

a saturated solution of Cu(IO3)2 was prepared in 0.0250 M KIO3. The copper concentration was found to be 0.00072 M. Set up an ICE chart and calculate the equilibrium (IO3-) and the Ksp

Answer 1

Answer:

$\large \boxed{\text{0.0264 mol/L; }5.0 \times 10^{-7}}$

Explanation:

1. The ICE table

$\begin{array}{rccccc} &\text{Cu}\text{(IO}_{3})_{2} & \rightleftharpoons &\text{Cu}^{2+}&+ & 2\text{IO}_{3}^{-}\\\\\text{I:}& & & 0 & & 0.0250\\\text{C:}& & & +x & & x \\\text{E:}& & & x & &0.0250 + 2x\\\end{array}$

2. Concentration of IO₃⁻

At equilibrium, [Cu²⁺] = 0.000 72 mol·L⁻¹, so x = 0.000 72.

The new ICE table becomes

$\begin{array}{rccccc} &\text{Cu}\text{(IO}_{3})_{2} & \rightleftharpoons &\text{Cu}^{2+}&+ & 2\text{IO}_{3}^{-}\\\\\text{I:}& & & 0 & & 0.0250\\\text{C:}& & & +0.00072 & & 0.00144 \\\text{E:}& & & 0.00072 & & \mathbf{0.0264}\\\end{array}\\\text{The equilibrium concentration of iodate ion is \large \boxed{\textbf{0.0264 mol/L}} }$

3. Ksp

$K_{\text{sp}} = [\text{Cu}^{2+}][\text{IO}_{3}^{-}]^{2} = 0.00072 \times 0.0264^{2} = \mathbf{5.0 \times 10^{-7}}$