Equilibrium Calculations: The “Small-X” Assumption

Many ICE table calculations will result in a quadratic (or higher) equation to solve. This can be time consuming – and in the case of cubic equations, may be impossible to do algebraically. To work around this problem, we can use the small-x approximation: a mathematical tool that will let us simplify some of the math, saving time.

Premise of the small-x approximation

In an ICE table, “x” represents the change (in M) during a reaction. For reactions with a small K value, we can expect this change to be very small. For very small changes, we can expect our initial values to be unchanged (or very nearly). for example:
$$\text{for [HI] = (1.00 M – x)} \qquad \text{and} \qquad \text {x = 0.003:}\\
[HI] = 1.00\, \text{M} – 0.003 \\
[HI] = 0.99_7 \, \text{M}\\
[HI] = 1.00\, \text{M}$$

The value of our change (x) was very small compared to the initial value, so our final value (within our sig fig limits) was the same as the initial value.

When we know the value of our change (x) will be small, we can make the assumption:
$$\mathbf{\text{initial value} \approx \text{final value}}$$

How do we know when we can use this approximation?

This approximation only works when “x” is small – if that’s not true, then we will introduce error into our calculations and make incorrect predictions about our equilibria. A general rule is that in most cases our experimental error (for example in the K value, or in the initial concentrations measured) is about 5%, so an “x” value less than about 5% of the initial value will not significantly increase the uncertainty into our calculations. There are two ways to predict if using values calculated with the small-x approximation are appropriate to use:

Way 1: Try it and see

One way to see if “x” is less than 5% of your initial value is just to try the calculations (using the approximation) and see. For example:
$$\text{[HI] (initial) = 1.00} \, \text{M. After solving an ICE table, [HI] (equilibrium) = 0.997} \, \text{M.}.\\
x = (1.00 – 0.997) = 0.003\\
%x = \frac{0.003}{1.00}\times 100% = 0.3%$$

“x” is 0.3% of our initial concentration – which is less than 5%, so in this case we are OK to continue the calculations using the equilibrium values calculated with the approximation.

In another calculation:
$$\text{[HI] (initial) = 1.00} \, \text{M. After solving an ICE table, [HI] (equilibrium) = 0.945} \, \text{M.}\\
x = (1.00 – 0.945) = 0.055\\
\% x = \frac{0.055}{1.00} = 5.5\%$$

5.5% is larger than 5%, so using the calculated value (0.945 M) may introduce extra uncertainty into our calculations (our answer may be wrong!). We should go back and re-calculate these values the long way, using the quadratic formula.

Way 2: Compare with K

We can save doing our calculations twice by using a quick comparison: if the initial concentration value is 1000x larger than K (or $\frac{1}{1000} \times$ smaller for reactions running in reverse), then our “change” value is likely to be small, and it is safe to use the approximation in solving the equilibrium calculations. For example:
$$\text{[HI] (initial) = 1.00} \, \text{M.}\; K = 1.03\times 10^{-5}\\
\frac{[initial]}{K} = \frac{1.00 }{ 1.03\times 10^{-5}} \approx 97000$$
This ratio is greater than 1000, so in this scenario we should be OK using the approximation.

It is always best to check that x < 5% of the initial value after using this approximation. You may find that some smaller ratios (where $100 \lt \frac{[initial]}{K} \lt 1000$) will still give acceptable values: in these edge cases, you can check the ratio and then verify your method by making sure your change is small (<5%). Ratios less than 100 will rarely work; it is best to just use the quadratic formula here.
*An exception to the “1000” rule is when you are working with an acid-base buffer; due to the specific nature of these reactions, a smaller ratio will work more often.

Using the small-x approximation in an ICE table

What are the concentrations at equilibrium of a 0.15 M solution of HCN? $$HCN\;(aq)\rightleftharpoons H^+\;(aq)+CN^-\;(aq)\qquad K_c=4.9×10^{-10}$$


Let’s check if the small-x approximation can be used: $$[HCN]_{initial} = 0.15\; \text{M} \quad K = 4.9×10^{-10}\\ \frac{0.15}{4.9×10^{-10}} = 3.0\times 10^{8} \gg 1000$$ We can assume that our change (x) will be small compared to this initial value (0.15 M).

Then, we build the ICE table:

  $HCN$ $\rightleftharpoons $ $H^+$ + $CN^-$
Initial Concentration 0.15   0   0
Change in Concentration -x   +x   +x
Equilibrium Concentration 0.15-x
$\approx $ 0.15
  x   x

The small-x approximation has been set up in the “Equilibrium” row for HCN. We can proceed with solving our equilibrium expression for x: $$K_C = \frac{[H^+][CN^-]}{[HCN]} \\ 4.9×10^{-10} = \frac{[x][x]}{[0.15]} \\ x^2 = 7.3_5 \times 10^{-11} \\ x = 8.5_{7321} \times 10^{-6}$$ Calculation check: Compare the value of x to our initial value to make sure the approximation was good: $$\frac{x}{[HCN]_{initial}} = \frac{8.5_{7321} \times 10^{-6}}{0.15} = 0.0057\% \lt 5%$$ So our assumption is good. We can continue with finding the equilibrium concentrations: $$\mathbf{[HCN] = 0.15 – x = 0.14_{999} \approx 0.15\\ [H^+] = [CN^-] = x = 8.6\times 10^{-6}}$$

As a check, you can repeat this calculation using the quadratic formula; your answers will be the same within significant figure limits.

Check Your Learning

What are the equilibrium concentrations in a 0.25 M NH3 solution? $$NH_3\;(aq) + H_2O\;(l) ⇌ NH_4^+\;(aq) + OH^−\;(aq)\qquad K_c=1.8×10^{−5}$$


Our inital concentration >> K (>1000x), so we can use the small-x approximation.

  $NH_3$ + $H_2 O\; (l)$ $\rightleftharpoons $ $NH_4 ^+$ + $OH^-$
Initial Concentration 0.25 M   $\require{cancel} \cancel{\;\;\;\;}$   0   0
Change in Concentration -x   -x   +x   +x
Equilibrium Concentration 0.25 -x
$\approx $ 0.25 M
  $\require{cancel} \cancel{\;\;\;\;}$   x   x
$$K_C = \frac{[NH_4 ^+][OH^-]}{[NH_3]}\\ 1.8×10^{−5} = \frac{[x][x]}{[0.25]}\\ x^2 = 4.5\times 10^{-6}\\ x = 2.1_{2132} \times 10^{-3}$$ Check the approximation: $$\frac{x}{[NH_3]_{initial}} = \frac{2.1_{2132} \times 10^{-3}}{0.25}=0.85\% \lt 5\% \quad ✓$$ Final concentrations: $$\mathbf{[NH_3] = 0.25 – x = 0.24_{7878}\; \text{M} = 0.25\; \text{M}\\ [NH_4 ^+] = [OH^-] = x = 2.1_{2132} \times 10^{-3}\; \text{M} = 2.1 \times 10^{-3}\; \text{M}}$$