Answer
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Hint: Henderson Hassel Balch equation describes relationship between $pH$ of blood and its components.it is also helpful in determining $pH$ of buffer solution. Amount of acid and base required to prepare a buffer solution can be determined using this equation.
Complete answer:
-By knowing pKa, pKb, equilibrium constant, buffer of known $pH$ can be prepared.
-To prepare acidic buffer, weak acid and its salt with strong base is used.
-weak acid ionizes in water as follows:
\[HA+{{H}_{2}}O\leftrightarrows{{H}_{3}}{{O}^{+}}+{{A}^{-}}\]
For which equilibrium constant can be written as follows:
\[Ka=\dfrac{[{{H}_{3}}{{O}^{+}}][{{A}^{-}}]}{[HA]}\]
Rearranging the equation gives
${{[{{H}_{3}}O]}^{+}}=\dfrac{Ka\text{ }[HA]}{[{{A}^{-}}]}$
Taking log on both sides and arranging terms we get,
$pH=pKa+\log \dfrac{[{{A}^{-}}]}{[HA]}$
This equation is known as Henderson-Hasselbalch equation.
The quantity $\dfrac{[{{A}^{-}}]}{[HA]}$ is the ratio of concentration of anion of acid and acid present in the mixture. Acid is a weak acid and it ionizes to very little extent and concentration of [HA] is negligibly different from concentration of acid taken from buffer.
Most of anion comes from ionization of salt or acid. Concentration of anion will be negligibly different from concentration of salt.
Equation can be written as
$pH=pKa+\log \dfrac{[salt]}{[acid]}$
So, to determine which acid or base should be selected to prepare a buffer of required $pH$.To determine the amount of acid and base required, to calculate pKa, pKb and to relate the bicarbonate buffer system of blood Henderson Hassel Balch equation is used.
Note:
When concentration of anion is equal to concentration of acid, then $pH$ is equal to pKa because the value of log 1 is zero. So, for preparing a buffer solution of required $pH$,acid whose pKa is close can be selected. In our body, there are various buffer systems which are in equilibrium. The Henderson HasselBalch equation is used to relate \[pH\] of the bicarbonate system of blood.
Complete answer:
-By knowing pKa, pKb, equilibrium constant, buffer of known $pH$ can be prepared.
-To prepare acidic buffer, weak acid and its salt with strong base is used.
-weak acid ionizes in water as follows:
\[HA+{{H}_{2}}O\leftrightarrows{{H}_{3}}{{O}^{+}}+{{A}^{-}}\]
For which equilibrium constant can be written as follows:
\[Ka=\dfrac{[{{H}_{3}}{{O}^{+}}][{{A}^{-}}]}{[HA]}\]
Rearranging the equation gives
${{[{{H}_{3}}O]}^{+}}=\dfrac{Ka\text{ }[HA]}{[{{A}^{-}}]}$
Taking log on both sides and arranging terms we get,
$pH=pKa+\log \dfrac{[{{A}^{-}}]}{[HA]}$
This equation is known as Henderson-Hasselbalch equation.
The quantity $\dfrac{[{{A}^{-}}]}{[HA]}$ is the ratio of concentration of anion of acid and acid present in the mixture. Acid is a weak acid and it ionizes to very little extent and concentration of [HA] is negligibly different from concentration of acid taken from buffer.
Most of anion comes from ionization of salt or acid. Concentration of anion will be negligibly different from concentration of salt.
Equation can be written as
$pH=pKa+\log \dfrac{[salt]}{[acid]}$
So, to determine which acid or base should be selected to prepare a buffer of required $pH$.To determine the amount of acid and base required, to calculate pKa, pKb and to relate the bicarbonate buffer system of blood Henderson Hassel Balch equation is used.
Note:
When concentration of anion is equal to concentration of acid, then $pH$ is equal to pKa because the value of log 1 is zero. So, for preparing a buffer solution of required $pH$,acid whose pKa is close can be selected. In our body, there are various buffer systems which are in equilibrium. The Henderson HasselBalch equation is used to relate \[pH\] of the bicarbonate system of blood.
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