Question

An environmental chemist needs a carbonate buffer of pH to study the effect of the acidification of limestone rich soils. How many grams of $N{{a}_{2}}C{{O}_{3}}$ must be added to \[1.5L\] of freshly prepared \[0.2{ }M\] $NaHC{{O}_{3}}$ to make the buffer? for ${{H}_{2}}C{{O}_{3}}$ , ${{K}_{{{a}_{1}}}}=4.7 \times {{10}^{-7}};{{K}_{{{a}_{2}}}}=4.7 \times {{10}^{-11}}$

Answer 1

Hint: Buffer solution is the solution which is formed by the mixture containing a weak acid and the conjugate base of weak acid or a weak base and the conjugate acid of the weak base.we have to first write the buffer reaction and then we can calculate rate constant through which we can calculate the concentration of reactant or product.

Complete step by step answer:
As we know buffer solutions are the solution formed by the mixing of weak acid or base and their conjugate acid or base. They can resist the change in pH when a small amount of dilution or addicting small acid or base.
These are of three types:
1) Acidic buffer formed when weak acid mix with its conjugate base
2) Basic buffer formed when a weak base combines with its conjugate acid.
3) And the third type is an inorganic buffer , which is formed when boric acid mixes with borax.

Mechanism of buffering, a salt is completely ionized and weak acid is partially ionized; the reaction can be chemically expressed as,
\[C{{H}_{3}}COONa\to C{{H}_{3}}CO{{O}^{-}}+N{{a}^{+}}\]
$C{{H}_{3}}COOH\to C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}$
On adding small amount of acid the proton of acid released will be removed by acetate ion and thus again forming buffer solution
In this way they resist change in pH via maintaining hydrogen ion concentration.
Now, com to the solution part , sodium bicarbonate is used to make buffer sodium carbonate is produce as a product reaction may be given as , $NaHC{{O}_{3}}+{{H}_{2}}O\to N{{a}_{2}}C{{O}_{3}}+{{H}_{3}}{{O}^{+}}$ Then ${{K}_{a}}$ may be written as - ${{K}_{a}}=\left[ {{H}_{3}}{{O}^{+}} \right]\left[ C{{O}_{3}}^{2-} \right]/\left[ HC{{O}_{3}}^{-} \right]$
THEN, after rearranging the equation the value of concentration of bicarbonate and acid dissociation constant on the left hand side of the equation, and the value of concentration of hydronium ion as well as carbonate ion, would remain on the right side of the equation. So, after substituting these values we get,
$=1\times {{10}^{-10}} \times 4.7 \times {{10}^{-11}}=\left[ C{{O}_{3}}^{2-} \right]\left( 1\times {{10}^{-10}} \right) \times 0.2$
$\Rightarrow \left[ C{{O}_{3}}^{2-} \right]=0.099Mol$
$\therefore $ then we will calculate the number of mole of sodium carbonate = $n=C \times V$ where $C$ is concentration , $V$ is volume
We know that the volume was given to us as \[1.5L\] and the concentration as we calculated came out to be, $0.099$, so after substituting the values we get,
$n=0.99\times 1.5$
$\Rightarrow n=0.14$
Then amount of sodium carbonate required to prepare \[0.2{ }M\] of sodium bicarbonate = $0.14 \times 106$ $=15g$ {$106=$ molecular mass of sodium carbonate}

Note: The pH range of the buffer solution belongs from \[1-10\].
-Human blood also contains a buffer of carbonic acid and bicarbonate anion in order to maintain the pH between \[7.35\] to \[7.45\].