ICE+problems

Equilibrium and ICE
The acronym I.C.E. stands for three conditions during a chemical reaction:
 * 1) **Initial (I)** conditions: the pre-reaction set-up - what reactants and products are present and in what concentration(s);
 * 2) **Changes (C)** that occur during the reaction (think reaction kinetics) - loss of reactants, gain in products;
 * 3) **Equilibrium (E)** conditions: the reaction is in a steady state and equilibrium is reached.

ICE problems are used for calculations of equilibrium-condition concentrations, when only the starting concentrations are known.

For example, what would the equilibrium concentrations of the reactants and products be in a 1.0 M solution of hydrofluoric acid (HF)?

HF is a weak acid, and would undergo hydrolysis:

HF (aq) + H 2 O (l) <--> H 3 O + (aq) + F - (aq)

The Ka value for HF is 6.8 x 10 -4.

Its equilibrium constant expression is as follows:

Ka = [H 3 O + ][F - ]/[HF]

Because its Ka value is so small, HF is only partially dissociated, so the molar concentrations of the products is not simply found using mole ratio. Here is where an ICE chart comes in handy. Set one up right under the chemical equation:


 * || HF (aq) || + || H 2 O (l) || <--> || H 3 O + (aq) || + || F - (aq) ||
 * Initial (I): ||  ||   ||   ||   ||   ||   ||   ||
 * Change (C): ||  ||   ||   ||   ||   ||   ||   ||
 * Equilibrium (E): ||  ||   ||   ||   ||   ||   ||   ||

Since water is a liquid, it is ignored:


 * || HF (aq) || + || H 2 O (l) || <--> || H 3 O + (aq) || + || F - (aq) ||
 * Initial (I): ||  ||   || -- ||   ||   ||   ||   ||
 * Change (C): ||  ||   || -- ||   ||   ||   ||   ||
 * Equilibrium (E): ||  ||   || -- ||   ||   ||   ||   ||

Now insert the initial conditions. There is 1.0 M HF and because the reaction has yet to start, there are no reactants:


 * || HF (aq) || + || H 2 O (l) || <--> || H 3 O + (aq) || + || F - (aq) ||
 * Initial (I): || 1.0 M ||  || -- ||   || 0 ||   || 0 ||
 * Change (C): ||  ||   || -- ||   ||   ||   ||   ||
 * Equilibrium (E): ||  ||   || -- ||   ||   ||   ||   ||

As the reaction proceeds, a quantity of HF is used up, and is converted according to mole ratio into the products. This unknown quantity is represented by "x":


 * || HF (aq) || + || H 2 O (l) || <--> || H 3 O + (aq) || + || F - (aq) ||
 * Initial (I): || 1.0 M ||  || -- ||   || 0 ||   || 0 ||
 * Change (C): || - x ||  || -- ||   || + x ||   || + x ||
 * Equilibrium (E): ||  ||   || -- ||   ||   ||   ||   ||

Finally, the equilibrium position is reached. This is simply the sum of the I and C rows:


 * || HF (aq) || + || H 2 O (l) || <--> || H 3 O + (aq) || + || F - (aq) ||
 * Initial (I): || 1.0 M ||  || -- ||   || 0 ||   || 0 ||
 * Change (C): || - x ||  || -- ||   || + x ||   || + x ||
 * Equilibrium (E): || 1.0 - x ||  || -- ||   || x ||   || x ||

Now use the E row by inserting its values into the equilibrium constant expression.


 * || HF (aq) || + || H 2 O (l) || <--> || H 3 O + (aq) || + || F - (aq) ||
 * Equilibrium (E): || 1.0 - x ||  || -- ||   || x ||   || x ||

Ka = 6.8 x 10 -4 = [H 3 O + ][F - ]/[HF] = (x)(x)/(1.0 - x) 6.8 x 10 -4 = x 2 /(1.0 - x)

From here, it can get tricky. You may have to simplify and use a quadratic equation to solve for x. Typically, though, x is so small that it will not have a significant effect on the initial concentration of HF, and it is assumed to be negligible. Therefore:

6.8 x 10 -4 = x 2 /1.0

and solving for x yields:

x = 0.026 M

Now refer to the ICE chart; x is the value for the molar concentrations of H 3 O + and F -, so therefore:

[H 3 O + ] = 0.026 M and [F - ] = 0.026 M.

The molar concentration of HF at equilibrium is 1.0 - x, so [HF] = 1.0 - 0.026 = 0.97 M.

From here, one could calculate the pH of this acid, etc.

Here is a video on ICE:

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