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**Hint:**We know that 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 answer:**

The solution is of two types named as an 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 pH buffer has a base 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.

$ H{{A}_{\left( aq \right)}}+OH_{\left( aq \right)}^{-}\to N{{a}^{+}}A_{\left( aq \right)}^{-}+{{H}_{2}}{{O}_{\left( l \right)}} $ So moles of $ HA=moles\text{ }of\text{ }{{A}^{-}}=0.1M\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 the buffer. The pH of a buffer solution of weak acid and conjugate base pair is given by the Henderson - Hasselbalch equation. It is represented as: $ pH=pKa+log\dfrac{\left[ {{A}^{-}} \right]}{\left[ HA \right]} $

Given the final pH of the solution is $ 5.5 $ and the $ pKa $ of the acid is $ 5. $

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

Thus, now by substituting the values in above equation;

$ 5.5=5+\log \left( \dfrac{0.01+x}{0.01-x} \right)\Rightarrow \log \left( \dfrac{0.01+x}{0.01-x} \right)=0.5 $

On further solving we get;

$ \dfrac{0.01+x}{0.01-x}=anti\log (0.5)\Rightarrow \dfrac{0.01+x}{0.01-x}=3.16 $

From above we get; $ 0.01+x=0.0316-3.16x $

Thus, we get; $ 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 given as;

$ =0.00519\times 40=0.210=2.10\times {{10}^{-1}}g. $

**Therefore, the correct answer is option A.**

**Note:**

Remember that buffer solution of weak acid and its salt. The Henderson Hasselbalch equation also applies for a weak base and salt 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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