# An acetic acid and sodium acetate buffer has pH = 4.76. The ratio of concentrations of [OAc]/[HOAc] is: pka of acetic acid = 4.76

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An acetic acid and sodium acetate buffer has pH = 4.76. The ratio of concentrations of [OAc]/[HOAc] is: pka of acetic acid = 4.76
1. 6 : 1
2. 4 : 3
3. 1 : 1
4. 4 : 1

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Correct Answer - Option 3 : 1 : 1

Concept:

• Buffer solution: A buffer solution contains either a weak acid and its salt with a strong base or a weak base and its salt with strong acid. Buffers resist a change in pH when acids or bases are added to them.

• Buffer can be classified into two types;

• Acidic Buffer: The buffer is formed from the solution of a weak acid and its salt of a strong base. acid. Acidic buffers generally have a pH of less than 7. Example of the acidic buffer is CH3COOH and CH3COONa.

• Basic Buffer: The buffer is formed from the solution of a weak base and its salt with a strong acid. basic buffers have a pH of more than 7. An example of the basic buffer is NH4OH and NH4Cl.

• The pH for a buffer solution is given by:
• For buffer solution,

${\rm{pH}} = {\rm{pKa}} + \log \left[ {\frac{{{\rm{salt}}}}{{{\rm{acid}}}}} \right]$

• This equation is known as Henderson - Hasselbalch equation and [ Salt ] is the concentration of the salt in buffer and [ Acid ] is the concentration of the acid in the buffer.
• pKa is the pK value of acid.
• When the concentration of the acid and the salt are equal, the pH of the buffer solution is equal to the pKa value of the acid.

Explanation:

• An acetic acid and sodium acetate buffer is an acidic buffer.
• The pH of the buffer is 5.36.
• pKa of the acid is given which is 4.76.
• From  Henderson - Hasselbalch equation, we know,

${\rm{pH}} = {\rm{pKa}} + \log \left[ {\frac{{{\rm{salt}}}}{{{\rm{acid}}}}} \right]$

Substituting the given values, we get:
${\rm{4.76}} = {\rm{4.76}} + \log \left[ {\frac{{{\rm{salt}}}}{{{\rm{acid}}}}} \right]$

or, ${\rm{4.76}} - {\rm{4.76}} = \log \left[ {\frac{{{\rm{salt}}}}{{{\rm{acid}}}}} \right] =0$

Hence, [salt] = [acid] as log 1 = 0.

Hence, the ratio, of concentrations of [OAc]/[HOAc] is 1:1.

• The buffer is at an equivalent point here.