Buffers are the solution that resists the change in pH. 1.39 ml of 5.90 M sodium hydroxide must be added to raise the pH to 5.75.
<h3>What is Henderson–Hasselbalch equation?</h3>
Henderson–Hasselbalch equation is used to determine the pH of the buffer by the equilibrium concentration of the acid and the conjugate base.
Moles of acetic acid are calculated as:

Moles of sodium acetate are calculated as:

The total number of moles is = 0.02352
Now, using the Henderson–Hasselbalch equation:
![\rm pH = \rm pKa + log \dfrac{[Acetate]}{[Acetic\; acid]}](https://tex.z-dn.net/?f=%5Crm%20pH%20%3D%20%5Crm%20pKa%20%2B%20log%20%5Cdfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3B%20acid%5D%7D)
Given,
pH = 5.75
pKa = 4.74
Substituting values in the above equation:
![\begin{aligned} 5.75 &= 4.74 +\rm log \rm \dfrac{[Acetate]}{[Acetic\; acid]}\\\\10.2329 &= \rm \dfrac{[Acetate]}{[Acetic\; acid]}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%205.75%20%26%3D%204.74%20%2B%5Crm%20%20log%20%5Crm%20%5Cdfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3B%20acid%5D%7D%5C%5C%5C%5C10.2329%20%26%3D%20%5Crm%20%5Cdfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3B%20acid%5D%7D%5Cend%7Baligned%7D)
Solving further:
![\begin{aligned} 10.2329 &= 0.02352 - \rm \dfrac{[Acetic\; acid]}{[Acetic\; acid]}\\\\11.2329\; \rm [Acetic \; acid] &= 0.02352 \\\\&= 0.002093\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%2010.2329%20%26%3D%200.02352%20-%20%5Crm%20%5Cdfrac%7B%5BAcetic%5C%3B%20acid%5D%7D%7B%5BAcetic%5C%3B%20acid%5D%7D%5C%5C%5C%5C11.2329%5C%3B%20%5Crm%20%5BAcetic%20%5C%3B%20acid%5D%20%26%3D%200.02352%20%5C%5C%5C%5C%26%3D%200.002093%5Cend%7Baligned%7D)
Moles of acetate is calculated as:
0.02352 moles - 0.002093 moles = 0.021427 moles
Initial moles of acetate were 0.0132, added moles of acetate by the addition of the sodium hydroxide resulted in, 0.021427 - 0.0132 = 0.00822 moles of sodium hydroxide.
Volume is calculated as:

Therefore, 1.39 mL of 5.90M of sodium hydroxide is added.
Learn more about Henderson–Hasselbalch equation here:
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