Question

An environmental chemist needs a carbonate buffer of pH 10 to study the effects of the acidification of limestone rich soils. How many grams of $ {\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} $ must be added to $ 1.5 $ Litre of freshly prepared $ 0.2 $ Molar $ {\text{NaHC}}{{\text{O}}_{\text{3}}} $ to make the buffer? For $ {{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} $ $ \left( {{K_{a1}} = 4.7 \times {{10}^{ - 7}}} \right) $ And $ \left( {{K_{a2}} = 4.7 \times {{10}^{ - 11}}} \right) $ .

Answer 1

Hint: Based on Lowry-Bronsted's acid-base theory, the conjugate acid is defined as the compound that is formed when the acid donates a proton or hydrogen ion to the base. Alternatively, the conjugate base is the leftover after an acid has donated its proton. We shall find the pKa of the solutions given and use the Henderson-Hasselbalch Equation to calculate the moles of $ {\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} $ required.
Formula Used: $ pH = p{K_a} - \log \left({\dfrac{{\left[{HA} \right]}}{{\left[{{A^ - }}\right]}}} \right) $
Where HA is the acid and $ {A^ - } $ is the conjugate acid.

Complete step by step Solution
As given in the question, the two dissociation constants for carbonic acid are $ \left( {{K_{a1}} = 4.7 \times {{10}^{ - 7}}} \right) $ and $ \left( {{K_{a2}} = 4.7 \times {{10}^{ - 11}}} \right) $ the acid dissociates in water as per the following reactions:
 $ {{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \rightleftarrows {\text{HC}}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{ + }}{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}} $ , and
 $ {\text{HC}}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \rightleftarrows {\text{C}}{{\text{O}}_{\text{3}}}^{{\text{2 - }}} + {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}} $
For these two reactions, the of the reaction medium = $ {\text{p}}{{\text{K}}_{{\text{a1}}}} = 7 - \log 4.7 $ and
 $ {\text{p}}{{\text{K}}_{{\text{a2}}}} = 11 - \log 4.7 $
Now, as per the Henderson- Hasselbalch Equation, the pH of the solution can be given by,
 $ {\text{pH = p}}{{\text{K}}_{\text{a}}}{\text{ + log}}\left( {\dfrac{{{{\text{n}}_{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}}}{{{\text{0}}{\text{.3}}}}} \right) $ , putting the value of $ {\text{p}}{{\text{K}}_{{\text{a2}}}} $ in the above equation, we get,
 $ {\text{10 = 11 - log 4}}{\text{.7 - log 0}}{\text{.3 + log }}{{\text{n}}_{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}} $
Or, $ \log \left( {\dfrac{{4.7 \times 0.3}}{{{{\text{n}}_{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}}}} \right) = 1 $ , removing the logarithm from both sides of the equation $ \left( {{\text{log 10 = 1}}} \right) $ we get,
 $ \left( {\dfrac{{4.7 \times 0.3}}{{10}}} \right) = {{\text{n}}_{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}} $ , or, $ {{\text{n}}_{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}} = 0.141 $
Now, the molecular weight of sodium carbonate = $ \left[ {\left( {23 \times 2} \right) + 12 + \left( {16 \times 3} \right)} \right] = 106 $ grams
Therefore according to the definition of mole, 1 mole of sodium carbonate = 106 grams.
So, $ 0.141 $ moles of sodium carbonate = $ 0.141 \times 106 $ grams = $ 14.946 $ grams.
Hence, grams of $ {\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} $ must be added to $ 1.5 $ Litre of freshly prepared $ 0.2 $ Molar $ {\text{NaHC}}{{\text{O}}_{\text{3}}} $ to make the buffer.

Notes
A buffer solution has different applications. They are added to any substance to keep its pH unchanged or even if it is changing then also the span of change is very little. An excellent example of the buffer solution is the human blood. The main reason for the resistance in the pH change of a buffer solution is that there exists an equilibrium between the conjugate acid and the base.