Answer
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Hint: The Henderson-Hasselbalch equation is a mathematical equation which connects the pH of the solution and the $p{K_a}$ which is equal to the $- \log {K_a}$. The ${K_a}$ is the acid dissociation constant of weak acid and its conjugate base.
Complete step by step answer:
The equation which relates the pH of an aqueous solution of an acid to the acid dissociation constant of the acid is described as the Henderson-Hasselbalch equation.
The equation is given as shown below.
$pH = p{K_a} + \log \left( {\dfrac{{[Conjugate\;base]}}{{[weak\;acid]}}} \right)$
The $p{K_a}$ value is equal to the negative logarithm of acid dissociation constant of the weak acid. It measures the strength of the acid's solution. The weak acid has $p{K_a}$ value ranging from 2-12 in water.
It is given as shown below.
$p{K_a} = - \log \left[ {{K_a}} \right]$
Where,
${K_a}$ is the acid dissociation constant of the weak acid.
The reaction of the weak acid-conjugate base buffer is shown below.
$HA(aq) + {H_3}O(l) \rightleftarrows {H_3}{O^ + }(aq) + {A^ - }(aq)$
The pH of the solution is given as shown below.
$pH = p{K_a} + \log \left( {\dfrac{{[{A^ - }]}}{{[HA]}}} \right)$
Now, we need to determine the ratio which exists between the concentration of the conjugate base, ${A^ - }$ and the concentration of the weak acid HA, add log on one side of the equation.
$\log \left( {\dfrac{{[{A^ - }]}}{{[HA]}}} \right) = pH - p{K_a}$
If x is equal to y,
${10^x} = {10^y}$
The above equation is equivalent to
${10^{\log \left( {\dfrac{{[{A^ - }]}}{{[HA]}}} \right)}} = {10^{pH - p{K_a}}}$
As we know,
${10^{\log 10(x)}} = x$
We get,
$\dfrac{{[{A^ - }]}}{{[HA]}} = {10^{pH - p{K_a}}}$
Note:
The Henderson-Hasselbalch equation is useful for determining the pH of the buffer solution and also determining the equilibrium pH in an acid-base reaction. The equation can also be used to determine the amount of acid and conjugate base used to prepare the buffer solution of a particular pH.
Complete step by step answer:
The equation which relates the pH of an aqueous solution of an acid to the acid dissociation constant of the acid is described as the Henderson-Hasselbalch equation.
The equation is given as shown below.
$pH = p{K_a} + \log \left( {\dfrac{{[Conjugate\;base]}}{{[weak\;acid]}}} \right)$
The $p{K_a}$ value is equal to the negative logarithm of acid dissociation constant of the weak acid. It measures the strength of the acid's solution. The weak acid has $p{K_a}$ value ranging from 2-12 in water.
It is given as shown below.
$p{K_a} = - \log \left[ {{K_a}} \right]$
Where,
${K_a}$ is the acid dissociation constant of the weak acid.
The reaction of the weak acid-conjugate base buffer is shown below.
$HA(aq) + {H_3}O(l) \rightleftarrows {H_3}{O^ + }(aq) + {A^ - }(aq)$
The pH of the solution is given as shown below.
$pH = p{K_a} + \log \left( {\dfrac{{[{A^ - }]}}{{[HA]}}} \right)$
Now, we need to determine the ratio which exists between the concentration of the conjugate base, ${A^ - }$ and the concentration of the weak acid HA, add log on one side of the equation.
$\log \left( {\dfrac{{[{A^ - }]}}{{[HA]}}} \right) = pH - p{K_a}$
If x is equal to y,
${10^x} = {10^y}$
The above equation is equivalent to
${10^{\log \left( {\dfrac{{[{A^ - }]}}{{[HA]}}} \right)}} = {10^{pH - p{K_a}}}$
As we know,
${10^{\log 10(x)}} = x$
We get,
$\dfrac{{[{A^ - }]}}{{[HA]}} = {10^{pH - p{K_a}}}$
Note:
The Henderson-Hasselbalch equation is useful for determining the pH of the buffer solution and also determining the equilibrium pH in an acid-base reaction. The equation can also be used to determine the amount of acid and conjugate base used to prepare the buffer solution of a particular pH.
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