E-StoreNumerical Approximations ? The good, bad and the ugly ?
Numerical Approximations – The good, bad and the ugly –
In this article, I will discuss about the application of Numerical Approximations as in XIth and XIIth class Standard level. I will discuss what a numerical approximation is, how it should be applied and how its application may come out to be good, bad or ugly. Please note that the below definitions/applications are not standards and I have tried to express the same based on my perception and experience. I will try to pull some examples from Chemistry where the concept can be applied. So, here we go –
What is a Numerical Approximation ?
It is a way to solve a numerical equation easily and with a relatively smaller and acceptable error (I will explain what I mean by this).
How to apply a numerical approximation ?
The rule to apply numerical approximation may be divided into 3 steps :
- Look for a pattern/condition/logic/motivation to apply the numerical approximation. Basically, your reason to apply it should be valid.
- Best part – Apply the approximation and solve the equation.
- Worst part – Check whether the approximation you applied gave a valid answer.
Examples :
1. Mole fraction of Urea in an aqueous solution is 0.010. Find the moles of urea in this solution prepared using 10 moles of water.
- We know, Mole fraction (
) = moles / total moles
= 0.01 = moles (urea) / [moles (urea) + moles (water)] where moles (water) = 10 = 0.01 = moles (urea) / [moles (urea) + 10]
If you see the above equation, there is no need to apply any approximation. You can simply use cross multiplication which will lead to a linear equation in moles (urea). We can save some marginal effort through approximations, however.
- Step – 1 :
- Think logically to apply numerical approximation (I mean you don’t need to apply it every time). You can see yourself that the mole fraction of urea is quite small which means that its moles are low as compared to the moles of water which further implies that its moles “may” be ignored as compared to that of water. The following may be a valid approximation :
moles (urea) + 10 
10(where
represents the approximation sign i.e. “nearly equal to”)
So, we have the logic to apply approximation.
Step – 2:
- Apply the approximation as follows :
= 0.01 = moles (Urea) / 10 moles (Urea) = 0.1Step – 3:
- This is a bit tricky and sometimes disappointing (because what you assumed/solved may not yield a good answer).
Here we will calculate the error which our approximation has generated. Logically, the error should be calculated as :
% error = [Mod (Actual answer – Approximate answer) / Actual answer ] * 100
Did you see that ? You need the actual answer of the numerical equation to calculate the error in your approximate answer. The problem is that we don’t want to calculate the actual answer as we want to solve the equation quickly. So, best way is to find the error in what we assumed vs what we got. Let’s have a look at our approximation again :
moles (urea) + 10 10
which means we assumed the sum to be “10” but it is coming out to be 0.1 + 10 = 10.1
So, % error in assumption = Mod(10.1 – 10)/10 * 100 % = 1%
We can say that our approximation (as above) will produce an error of 1%. Note that the acceptable range is generally considered to be 4 – 5 %. So, if your error is above the threshold, then you can’t apply the approximation to get an answer closer to the actual (exact) answer.
In this case, however, we can also calculate the Exact answer as :
= 0.01 = moles (urea) / [moles (urea) + 10] moles (urea) = 0.1/0.99
If you assumed that the approximations always work out to be good, you are mistaken. Let’s assume that the mole fraction value was 0.5 instead. In this case, if you use the above approach, you get two answers as :
Exact answer : moles (urea) = 10 Approximate answer : moles (urea) = 5
Did you see the difference ?
Let’s take another example :
2. What is the value of the degree of dissociation of a 0.1 M weak acid (HA) (dissociation constant = 10-5) solution ?
- The related equation for a weak acid dissociation is given by : Ka = c
/ (1 – 
)
Step – 1 : Acid is weak so its degree of dissociation will be small. Thus, 1 – 
1
Step – 2 : The equation becomes : Ka = c = 0.01
Step – 3 : % error = (1 – (1- ))/1 * 100 = 1% (acceptable)
I hope the basic idea of applying the approximation is clear. I will take more examples in Part – II of this article very soon.
Cheers
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