One liter of a buffer solution contains 0.004 mole of acetic acid (pKa= 4.76) and 0.4 mole of sodium acetate. The pHof the solution is __________. (Round off to two decimal places)Correct answer is between '6.66,6.86'. Can you explain this answer?

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Answer

Answer

Calculating pH of a Buffer Solution

In order to calculate the pH of a buffer solution, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

pH = the pH of the buffer solution

pKa = the acid dissociation constant of the weak acid component of the buffer

[A-] = the molar concentration of the conjugate base component of the buffer

[HA] = the molar concentration of the weak acid component of the buffer

Calculating the Concentration of the Conjugate Base

We can use the balanced chemical equation of the dissociation of acetic acid to find the concentration of the conjugate base (sodium acetate):

CH3COOH ⇌ CH3COO- + H+

For every mole of acetic acid that dissociates, one mole of acetate ion is produced.

Since the solution contains 0.004 moles of acetic acid, we know that the concentration of acetate ion is also 0.004 M.

Calculating the Concentration of the Weak Acid

We know that the total concentration of the buffer solution is 1 liter. This means that the concentration of the weak acid (acetic acid) can be calculated as follows:

[HA] = total concentration – [A-] = (1 – 0.4) = 0.6 M

Calculating pH

Now that we have the values for pKa, [A-], and [HA], we can plug them into the Henderson-Hasselbalch equation:

pH = 4.76 + log(0.4/0.6) ≈ 6.76

Conclusion

The pH of the buffer solution is approximately 6.76, which falls within the given range of 6.66 to 6.86.

Similar Chemistry Doubts

The measured freezing point depression for 0.1m aqueous acetic acid solution is 0.190C. Ka at this concentration will be (Kf = 1.86 K molal–1). [Give answer in multiple of 10–5 rounded up to onedecimal places] Correct answer is between '3.9,4.1'. Can you explain this answer?

What volume (in ml, rounded up to first decimal plac e) of 0.10 M sodium formate solution should be added to 50 ml of 0.05 M formic acid to produce a buffer solution of pH 4. [pKa for formic acid is 3.7] Correct answer is between '49.5,50.5'. Can you explain this answer?

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