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Why is the curve shaped like this?
The shape of the curve can be explained by the logarithmic nature of the pH scale. The pH value of a solution is defined as the negative decadic logarithm of the hydronium-ion concentration
$$\mathrm{pH} = -\log{c\mathrm{(H_3O^+)}}.$$
When 90 % of the acid is neutralised, the pH value is changed by one unit. When further 9 % are neutralised and overall 99 % acid is converted, the pH value changes by another unit. Next, only 0.9 % more acid needs to be converted, to change the pH value by another unit. Because of this logarithmic behaviour, the pH value does not change much at the beginning of the titration. As we get closer to the equivalence point, the pH value rises quickly until there is a jump around the equivalence point. Afterwards, the pH level rises because of the added base.
Here you can experiment with some parameters and how they change the titration curve.
Next, you need to know the concentration of your titrant, in this example the base, also in $\frac{\text{mol}}{\text{L}}$.
Finally, let's take a look at the concentration of the titrant. Try to make a prediction, how it affects the curve and check it afterwards in the same way as before.
Here you can track the titration by visualising the point in the titration curve at a given added volume of titrant. The slider can also be controlled with the arrow keys for more precise control, which is most important around the equivalence point.
As for the deprotonated acid $\mathrm{A^-}$, we need to consider the amount of substance at the beginning $n_0\mathrm{(A^-)}$ caused by the equilibrium shown above and additionally the amount of substance formed by the deprotonation of acid $n_{\mathrm{d}}\mathrm{(A^-)}$ with the added base
$$\mathrm{HA} + \mathrm{NaOH} \longleftrightarrow \mathrm{Na^+} + \mathrm{A^-} + \mathrm{H_2O}.$$
Consequently, we can write
$$n\mathrm{(A^-)} = n_0\mathrm{(A^-)} + n_{\mathrm{d}}\mathrm{(A^-)}.$$