
How do you determine $pH$ from $p{{K}_{a}}$?
Answer
548.1k+ 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, $pH = -\log \left[ HA \right]$
Recently Updated Pages
Why is there a time difference of about 5 hours between class 10 social science CBSE

In cricket, what is a "pink ball" primarily used for?

In cricket, what is the "new ball" phase?

In cricket, what is a "death over"?

What is the "Powerplay" in T20 cricket?

In cricket, what is a "super over"?

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

