# Question f7924

Sep 29, 2017

They react and form copper iodide(I) ($C u I$) and pottasium triodide($K {I}_{3}$).
Thank you for Gaya3's post.

#### Explanation:

I thought that reaction with copper sulfate($C u S {O}_{4}$) and potassium iodide($K I$) does'nt happen for two reasons, but it was wrong.

This is what I have written:

(1) Potassium cation (${K}^{+}$) is very stable. $K$ is one of alkali metals, which lies in the group 1 in the periodic table. In general,
alkali metal cations form no precipitation with any anions.

(2) Iodine anion (${I}^{-}$) is also very stable. $I$ is one of halogens,
in the group 17 in the periodic table. Halogen anions don't
precipitate expect with $A {g}^{+}$ and $P {b}^{2 +}$.

What I did not notice was the fact that ${I}^{-}$ has strong reducibility. Please see Gaya3's post for the chemical equation.

Sep 29, 2017

Yes!

#### Explanation:

Equation:

color(blue)(CuSO_4) + color(red)(KI) -> color(orange)(CuI) + color(violet)(K_2SO_4) + color(green)(KI3

After Balancing,

color(blue)(2CuSO_4 + color(red)(5KI -> color(orange)(2CuI + color(violet)(2K_2SO_4 + color(green)(KI_3#

Watch it happen here