
How do you determine $pH$ from $p{{K}_{a}}$?
Answer
575.1k+ views
Hint: The relation between $pH$ and $p{{K}_{a}}$ can be established using the Henderson and Hasselbalch equation.
- $p{{K}_{a}}$ is the measure of the strength of an acid especially used for weak acids.
Complete step by step answer:
In the question it is asked how we will determine the$pH$for an acid if the $p{{K}_{a}}$ given.
- First let’s see what $pH$ and $p{{K}_{a}}$ value means.
The $pH$ value gives the power of the hydrogen ion present in the aqueous solution i.e. the $pH$ value scale is used to determine whether the aqueous solution is acidic or basic in nature. $pH$ is inversely equal to the hydrogen concentration in the aqueous solution.
$pH=-\log \left[ {{H}^{+}} \right]$
- The $p{{K}_{a}}$ value is the measure to represent the strength of the weak acid, generally strength of the strong acid are represented by $pH$ values but representing the$pH$value for weak acid is awkward since the $pH$ value obtained will be have powers raised to ten which will make the calculations hectic.
$p{{K}_{a}}=-\log {{K}_{a}}$
- For solving the question given lets take the case of monoprotic acid i.e. an acid which has the ability to donate only one proton.The taken acid is a buffer solution i.e. a solution of weak acid and its conjugate base.
- Let's write the dissociation of the buffer taken,
$H{{A}_{(aq)}}\rightleftharpoons {{H}^{+}}_{\left( aq \right)}+{{A}^{-}}_{\left( aq \right)}$
- We have discussed above that we are dealing with the buffer hence it is a weak solution and the dissociation constant is represented as $p{{K}_{a}}$ and we have also written the equation above relating acid dissociation ( for strong acids) ${{K}_{a}}$ and $p{{K}_{a}}$
We know that equation that relates $p{{K}_{a}}$ and $pH$ value of a buffer is the Henderson-Hasselbalch equation which is ,
$pH=p{{K}_{a}}+\log \dfrac{\left[ {{A}^{-}} \right]}{\left[ HA \right]}$
- Here $\left[ {{A}^{-}} \right]$ is the concentration of the conjugate base and $\left[ HA \right]$ is the concentration of the weak acid.
If the concentrations and the $p{{K}_{a}}$ value is known then we could find the $pH$ value of the buffer using this equation.
Note: If we are not dealing with buffer solution and the given sample is a strong acid then we know that the strong acids dissociates completely in aqueous solution.For strong acids we take ${{K}_{a}}$ as dissociation constant.
- We know $p{{K}_{a}} = -\log {{K}_{a}}$ and $pH = -\log \left[ {{H}^{+}} \right]$
- Since in strong acids complete dissociation takes place, concentrations $\left[ HA \right]=\left[ {{H}^{+}} \right]$
Therefore, $pH = -\log \left[ HA \right]$
- $p{{K}_{a}}$ is the measure of the strength of an acid especially used for weak acids.
Complete step by step answer:
In the question it is asked how we will determine the$pH$for an acid if the $p{{K}_{a}}$ given.
- First let’s see what $pH$ and $p{{K}_{a}}$ value means.
The $pH$ value gives the power of the hydrogen ion present in the aqueous solution i.e. the $pH$ value scale is used to determine whether the aqueous solution is acidic or basic in nature. $pH$ is inversely equal to the hydrogen concentration in the aqueous solution.
$pH=-\log \left[ {{H}^{+}} \right]$
- The $p{{K}_{a}}$ value is the measure to represent the strength of the weak acid, generally strength of the strong acid are represented by $pH$ values but representing the$pH$value for weak acid is awkward since the $pH$ value obtained will be have powers raised to ten which will make the calculations hectic.
$p{{K}_{a}}=-\log {{K}_{a}}$
- For solving the question given lets take the case of monoprotic acid i.e. an acid which has the ability to donate only one proton.The taken acid is a buffer solution i.e. a solution of weak acid and its conjugate base.
- Let's write the dissociation of the buffer taken,
$H{{A}_{(aq)}}\rightleftharpoons {{H}^{+}}_{\left( aq \right)}+{{A}^{-}}_{\left( aq \right)}$
- We have discussed above that we are dealing with the buffer hence it is a weak solution and the dissociation constant is represented as $p{{K}_{a}}$ and we have also written the equation above relating acid dissociation ( for strong acids) ${{K}_{a}}$ and $p{{K}_{a}}$
We know that equation that relates $p{{K}_{a}}$ and $pH$ value of a buffer is the Henderson-Hasselbalch equation which is ,
$pH=p{{K}_{a}}+\log \dfrac{\left[ {{A}^{-}} \right]}{\left[ HA \right]}$
- Here $\left[ {{A}^{-}} \right]$ is the concentration of the conjugate base and $\left[ HA \right]$ is the concentration of the weak acid.
If the concentrations and the $p{{K}_{a}}$ value is known then we could find the $pH$ value of the buffer using this equation.
Note: If we are not dealing with buffer solution and the given sample is a strong acid then we know that the strong acids dissociates completely in aqueous solution.For strong acids we take ${{K}_{a}}$ as dissociation constant.
- We know $p{{K}_{a}} = -\log {{K}_{a}}$ and $pH = -\log \left[ {{H}^{+}} \right]$
- Since in strong acids complete dissociation takes place, concentrations $\left[ HA \right]=\left[ {{H}^{+}} \right]$
Therefore, $pH = -\log \left[ HA \right]$
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