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The problem we did during lecture involved finding the ph and ionization of acetic acid in .10M CH3COOH , with K=1.8x10^-5. I was bit confused why we weren't suppose to use the quadratic equation, I heard the professor saying to omit this step since its a weak acid?
When the Ka value of the acid is less than 10^-3, we can approximate the value of x by assuming the value of x being subtracted from 0.10 to be insignificant. Even if we subtract the value of x from 0.10, the value will still essentially be 0.10 because of how small x will be.
It is because the Kc is very small, that the change in negligible. I think the example that Dr. Lavelle gave was if you have a million dollars and you take for instance, a dollar away, the value is still very close to one million, so you can approximate it to one millon. After you do that, it is no longer a quadratic equation.
Because the K value is so small you can use the approximation method. In this case the x value will be so small that you can assume the value (0.1-x) will be approximately equal to 0.1. You can use this method whenever K is less than 10^-3.
The Ka is incredibly small (less than 10^-3), and so we don't need to use the quadratic equation because we assume the molar concentration of the original protonated molecule is going to be essentially unchanged at equilibrium. Though some of the molecule will be deprotonated, it is such a small number that when we use sig figs, the overall concentration will not budge. For example, if the concentration at equilibrium was 0.5-x and x=.0000001, it would not make sense to include the "-x" term in the calculation. It complicates beyond the degree of precision we are employing for our purposes.
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