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Burka [1]
2 years ago
7

How much 5. 90 m naoh must be added to 480. 0 ml of a buffer that is 0. 0215 m acetic acid and 0. 0275 m sodium acetate to raise

the ph to 5. 75?
Chemistry
1 answer:
AveGali [126]2 years ago
6 0

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:

0.480 \times 0.0215 = 0.01032

Moles of sodium acetate are calculated as:

0.480 \times 0.0275 = 0.0132

The total number of moles is = 0.02352

Now, using the Henderson–Hasselbalch equation:

\rm pH = \rm pKa + log \dfrac{[Acetate]}{[Acetic\; acid]}

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}

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}

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:

0.00822 \;\rm moles \times \dfrac{1\;\rm L}{5.90\;\rm M} = 0.00139 L

Therefore, 1.39 mL of 5.90M of sodium hydroxide is added.

Learn more about Henderson–Hasselbalch equation here:

brainly.com/question/19755735

#SPJ4

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