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? $$HCN\;(aq)\rightleftharpoons H^+\;(aq)+CN^-\;(aq)\qquad K_c=4.9×10^{-10}$$

**Solution**

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* NH_{3} solution?
$$NH_3\;(aq) + H_2O\;(l) ⇌ NH_4^+\;(aq) + OH^−\;(aq)\qquad K_c=1.8×10^{−5}$$

## Answer

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 |