Question

Consider a solution of a weak acid at a $ pH $ equal to its $ pKa. $ By how much would the $ pH $ change, and in which direction, if we added to this solution enough base to neutralize $ 10% $ of the total acid?

Answer 1

Hint: The pH is the main factor used to differentiate between the neutralization reaction and the buffer solution. In neutralization reaction, salt is formed by reaction acid with the base whereas the buffer solution is prepared by mixing weak acid and its conjugate base and vice versa.

Complete answer:
The neutralization reaction is defined as the reaction which takes place between the acid and the base which results in the formation of a salt by eliminating water as the by-product. The general neutralization reaction is given by; $ Acid+Base\to Salt+water.\text{ } $
The pH of the salt is usually $ 7 $ the buffer solution is a water solvent which is prepared by the mixture containing a weak acid and the conjugate base of the weak acid or a weak base and the conjugate acid of the weak base. The property of the buffer solution is that it resists any change in pH when any small amount of acid and base is added.
 $ pH=pKa+log\left( \dfrac{[conjugate\text{ }base]}{[weak\text{ }acid]} \right) $ As you can see here, equal amounts of weak acid and of conjugate base will make the long term equal to zero, and thus the pH equal to the $ pKa. $
Now you're adding a volume v of strong base. If $ 10% $ of the acid is neutralized, the number of moles you'll be left with is $ {{x}_{acid}}=x-\dfrac{10}{100}x=\dfrac{9}{10}x $
The number of moles of conjugate base will increase by the same amount, so you have
 $ {{x}_{base}}=x+\dfrac{10}{100}x=\dfrac{11}{10}x $ The new concentrations of weak acid and conjugate base will be
 $ [HA]=\left( \dfrac{9}{10}x \right)\left( \dfrac{1}{V+v} \right) $ and $ [{{A}^{-}}]=\left( \dfrac{11}{10}x \right)\left( \dfrac{1}{V+v} \right) $
Putting this into the $ H-H $ equation to get;
 $ pH=pKa+\log \left[ \dfrac{\left( \dfrac{11}{10}x \right)\left( \dfrac{1}{V+v} \right)}{\left( \dfrac{9}{10}x \right)\left( \dfrac{1}{V+v} \right)} \right] $
 $ pH=pKa+\log \left( \dfrac{11}{9} \right) $
 $ \Rightarrow pH=pKa+0.087 $
The pH of the solution will thus increase by about $ ~0.09. $.

Note:
You must know that the buffer solution can be formed by adding a strong base to the solution of weak acid to neutralize half of the weak acid to its conjugate base example converting half of the concentration of acetic acid to its conjugate base acetate ion.