Iodine is sparingly soluble in pure water. However, it does $`$dissolve$\text{'}$ in solutions containing excess iodide ion because of the following reaction:

I$-($aq$)+$ I$2($aq$)=$ I$3-($aq$)$  K $=710.$

For each of the following cases calculate the equilibrium ratio of $[$I$3-]$ to $[$I$2].$

a$)6\mathrm{.}00\backslash$times $10-2$ mol of I$2$ is added to $1\mathrm{.}00$ L of $6\mathrm{.}00\backslash$times $10-1$ M KI solution.

b$)$The solution above is diluted to $10\mathrm{.}50$ L$.$

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To solve this problem we need to use the equilibrium constant expression for the reaction Iaq I2aq I... View the full answer