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A buffer is prepared by adding 300.0 mL of 2.0 M NaOH to 500.0 mL of...

Question

A buffer is prepared by adding 300.0 mL of 2.0 M NaOH to 500.0 mL of...

A buffer is prepared by adding 300.0 mL of 2.0 M NaOH to 500.0 mL of 2.0 M CH3COOH. What is the pH of the buffer?

Answers

Solution :-

Ka (acid dissociation constant) for CH3COOH = 1.8 × 10-5

pKa = -log10 ​​​​​​Ka = -log10 (1.8 × 10-5) = 4.744

Volume of NaOH = 300mL = 0.3 L

Molarity of NaOH = 2 M

Number of moles of NaOH = Molarity × Volume = (2 × 0.3) = 0.6 mole

Volume of CH​​​​​​​3COOH = 0.5 L

Molarity of CH3COOH = 2 M

Number of moles of CH3COOH = Molarity × Volume = (2 × 0.5) = 1 mole

On mixing the NaOH and CH​​​​​​​3COOH solutions, change in the volume and concentration occurs.

Now, total volume of solution formed on mixing = (0.3 + 0.5) = 0.8 L

New molarity of NaOH = moles / new volume = 0.6 / 0.8 = 0.75 M

New molarity of CH​​​​​​​​​​​​​​3COOH = moles / new volume = 1 / 0.8 = 1.25 M

Using Henderson-hasselbalch equation:-

pH = pKa + log10 [Base] / [Acid]

pH = 4.744 + log10 [0.75] / [1.25]

pH = 4.744 + (-0.222)

pH = 4.522

Hence, pH of the buffer solution = 4.522

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