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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.
What is Henderson–Hasselbalch equation?
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:
[tex]0.480 \times 0.0215 = 0.01032[/tex]
Moles of sodium acetate are calculated as:
[tex]0.480 \times 0.0275 = 0.0132[/tex]
The total number of moles is = 0.02352
Now, using the Henderson–Hasselbalch equation:
[tex]\rm pH = \rm pKa + log \dfrac{[Acetate]}{[Acetic\; acid]}[/tex]
Given,
pH = 5.75
pKa = 4.74
Substituting values in the above equation:
[tex]\begin{aligned} 5.75 &= 4.74 +\rm log \rm \dfrac{[Acetate]}{[Acetic\; acid]}\\\\10.2329 &= \rm \dfrac{[Acetate]}{[Acetic\; acid]}\end{aligned}[/tex]
Solving further:
[tex]\begin{aligned} 10.2329 &= 0.02352 - \rm \dfrac{[Acetic\; acid]}{[Acetic\; acid]}\\\\11.2329\; \rm [Acetic \; acid] &= 0.02352 \\\\&= 0.002093\end{aligned}[/tex]
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:
[tex]0.00822 \;\rm moles \times \dfrac{1\;\rm L}{5.90\;\rm M} = 0.00139 L[/tex]
Therefore, 1.39 mL of 5.90M of sodium hydroxide is added.
Learn more about Henderson–Hasselbalch equation here:
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