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**Hint:**A buffer solution is a solution which resists a change in the pH due to the addition of acid or base. The pH of a buffer solution is expressed in terms of the dissociation constant of the acid or and its conjugate base.

**Complete step by step answer:**

Buffer solution is of two types named as acidic buffer or a basic buffer. An Acid buffer has an acidic \[pH\] which is made of a weak acid and its salt with a strong base. Similarly a basic buffer has a basic \[pH\] made of a weak base and its salt with a strong acid.

The addition of sodium hydroxide to a solution reacts with the acid \[HA\] and produces a salt of sodium as \[N{a^ + }{A^ - }\] and water. The acid and the base \[NaOH\] react in \[1:1\] mole ratios.

\[HA\left( {aq} \right) + O{H^ - }\left( {aq} \right) \to N{a^ + }{A^ - }\left( {aq} \right) + {H_2}O\left( l \right)\]

Next we determine the number of moles of weak acid and the conjugate base present in the buffer before the addition of the strong base. The molarity of the acid is given as \[0.1M\] and the volume of the buffer solution is \[100mL\].

So moles of \[HA\] = moles of \[{A^ - }\] = \[0.1M \times 100mL = \dfrac{{0.1{\text{ }}moles}}{{1000mL}} \times 100mL = 0.01moles\].

Before determining the actual amount in grams of sodium hydroxide we calculate the number of moles of sodium hydroxide added to buffer solution. Let \[x\] be the number of moles of sodium hydroxide added to the buffer.

So at the end of the reaction between \[NaOH\] and \[HA\] , the resulting solution contains \[0.01 - x\] moles of \[HA\] where \[x\] moles of the acid is consumed. The moles of the conjugate base \[{A^ - }\] produced in the reaction is \[0.01 + x\] moles.

The \[pH\] of a buffer solution of weak acid and conjugate base pair is given by Henderson - Hasselbalch equation. It is represented as

\[pH = pKa + log\dfrac{{[A - ]}}{{[HA]}}\]

Given the final \[pH\] of the solution is \[5.5\] and the \[pKa\] of the acid is \[5\].

Thus \[pH = pKa + log\left[ {\dfrac{{\left( {0.01 + x} \right)moles}}{{\left( {0.01 - x} \right)moles}}} \right]\]

\[5.5 = 5 + log\left( {\dfrac{{0.01 + x}}{{0.01 - x}}} \right)\]

\[log\left( {\dfrac{{0.01 + x}}{{0.01 - x}}} \right) = 0.5\]

\[\dfrac{{0.01 + x}}{{0.01 - x}} = anti\log (0.5)\]

\[\dfrac{{0.01 + x}}{{0.01 - x}} = 3.16\]

\[0.01 + x = 0.0316 - 3.16x\]

\[x = \dfrac{{0.0316 - 0.01}}{{1 + 3.16}} = 0.00519moles.\]

The molar mass of sodium hydroxide is \[40g/mol\]. Thus the amount of sodium hydroxide is calculated as

$ = 0.00519moles \times 40g/mol = 0.21g$.

**Hence \[0.21g\] of solid \[NaOH\] must be added to a buffer solution to make the pH of solution \[5.5\].**

**Note:**This is an example of a buffer solution of weak acid and its salt. The Henderson Hasselbalch equation also applies for a weak base and salt buffer. Buffer system is very important in several biological processes to maintain the pH of the blood. The enzymatic activity in living systems is dependent on the pH of the blood.

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