ICE Tables - Chemistry LibreTexts

Solution

This equation describes a weak acid reaction in solution with water. The acid (HA) dissociates into its conjugate base (\(A^-\)) and protons (H3O+). Notice that water is a liquid, so its concentration is not relevant to these calculations.

STEP 1: Fill in the given concentrations

Reaction:	HA	A-	H3O+
I	0.150 M	0.000 M	0.000 M
C	?	?	?
E	?	?	?

• The contents of the leftmost column column are shortened for convenience.

STEP 2: Calculate the change concentrations by using a variable 'x'

Reaction:	HA	A-	H3O+
I	0.150 M	0.000 M	0.000 M
C	-x M	+x M	+x M
E	?	?	?

• The change in concentration is unknown, so the variable x is used to denote the change. x is the same for both products and reactants because equal stoichiometric amounts of A- and H3O+ are generated when HA dissociates in water.

STEP 3: Calculate the concentrations at equilibrium

Reaction:	HA	A-	H3O+
I	0.150 M	0.000 M	0.000 M
C	-x M	+x M	+x M
E	0.150 - x M	x M	x M

• To find the equilibrium amounts the I row and the C row are added. Use these values and Ka (the equilibrium constant for acids) to find the concentration x.

STEP 4: Use the ICE table to calculate concentrations with \(K_a\)

The expression for Ka is written by dividing the concentrations of the products by the concentrations of the reactants. Plugging in the values at equilibrium into the equation for Ka gives the following:

\[K_a = \dfrac{x^2}{0.150-x} = 1.6 \times 10^{-2} \nonumber\]

To find the concentration x, rearrange this equation to its quadratic form, and then use the quadratic formula to find x:

\[\begin{align*} (1.6 \times 10^{-2})({0.150-x}) &= {x^2} \\[4pt] x^2+(1.6 \times 10^{-2})x-(0.150)(1.6 \times 10^{-2}) &= 0 \end{align*}\]

This is the typical form for a quadratic equation:

\[Ax^{2}+Bx+C=0\nonumber \]

where, in this case:

• \(A = 1\)
• \(B = 1.6 \times 10^{-2}\)
• \(C =( -0.150)( 1.6 \times 10^{-2}) = -2.4 \times 10^{-3} \)

The quadratic formula gives two solutions (but only one physical solution) for x:

\[x = \dfrac{-B+\sqrt{B^2-4AC}}{2A}\nonumber \]

and

\[x = \dfrac{-B-\sqrt{B^2-4AC}}{2A}\nonumber \]

Intuition must be used in determining which solution is correct. If one gives a negative concentration, it can be eliminated, because negative concentrations are unphysical.

The x value can be used to calculate the equilibrium concentrations of each product and reactant by plugging it into the elements in the E row of the ice table.

[Solution: x = 0.0416, -0.0576. x = 0.0416 makes chemical sense and is therefore the correct answer.]