
Write the Henderson - Hasselbalch equation?
Answer
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Hint: For this problem, we have to study the concept of buffer solution as the equation is used to estimate the pH of a buffer solution. Also, we have to study the theory of the equation as well as its application.
Complete step by step solution:
-In the given question, we have to explain the Henderson - Hasselbalch equation in detail.
-There are two scientists which gave the equation i.e. Henderson gave the equation to calculate the pH of a buffer solution whereas Hasselbalch gave the logarithmic form of the equation.
-Now, as we know that a buffer solution consists of an acid and its conjugate base. So, the equation gives the relation between the pH of a solution, acid dissociation constant and the concentration of the species of the solution.
-For the relation, both scientists were given some assumptions that are:
1.The monobasic acid is used which dissociates into respective ions i.e.
$\text{HA }\rightleftharpoons \text{ }{{\text{H}}^{+}}\text{ + }{{\text{A}}^{-}}$ ……(1)
2.This assumption says that the self-ionisation of the water can be ignored and doesn't valid when the pH is more than 10.
3.Now, the thermodynamic constant K, for equation first can be written as:
$\text{K = }\dfrac{({{\text{H}}^{+}})({{\text{A}}^{-}})}{(\text{HA})}$ ….. (2)
-Here, the species present in the brackets signifies the concentration of the reactant and product.
-Now, when the logarithm of the equation (2) is taken then we get:
$\text{pH = p}{{\text{K}}_{\text{a }}}+\text{ lo}{{\text{g}}_{10}}\left( \dfrac{{{\text{A}}^{-}}}{\text{HA}} \right)$
Therefore, $\text{pH = p}{{\text{K}}_{\text{a }}}+\text{ lo}{{\text{g}}_{10}}\left( \dfrac{{{\text{A}}^{-}}}{\text{HA}} \right)$ is a Henderson - Hasselbalch equation.
Note: The main application of this equation is to determine the pH of a solution which consists of an acid and its conjugate base. In the equation, $\text{p}{{\text{K}}_{\text{a }}}$is the acid dissociation constant.
Complete step by step solution:
-In the given question, we have to explain the Henderson - Hasselbalch equation in detail.
-There are two scientists which gave the equation i.e. Henderson gave the equation to calculate the pH of a buffer solution whereas Hasselbalch gave the logarithmic form of the equation.
-Now, as we know that a buffer solution consists of an acid and its conjugate base. So, the equation gives the relation between the pH of a solution, acid dissociation constant and the concentration of the species of the solution.
-For the relation, both scientists were given some assumptions that are:
1.The monobasic acid is used which dissociates into respective ions i.e.
$\text{HA }\rightleftharpoons \text{ }{{\text{H}}^{+}}\text{ + }{{\text{A}}^{-}}$ ……(1)
2.This assumption says that the self-ionisation of the water can be ignored and doesn't valid when the pH is more than 10.
3.Now, the thermodynamic constant K, for equation first can be written as:
$\text{K = }\dfrac{({{\text{H}}^{+}})({{\text{A}}^{-}})}{(\text{HA})}$ ….. (2)
-Here, the species present in the brackets signifies the concentration of the reactant and product.
-Now, when the logarithm of the equation (2) is taken then we get:
$\text{pH = p}{{\text{K}}_{\text{a }}}+\text{ lo}{{\text{g}}_{10}}\left( \dfrac{{{\text{A}}^{-}}}{\text{HA}} \right)$
Therefore, $\text{pH = p}{{\text{K}}_{\text{a }}}+\text{ lo}{{\text{g}}_{10}}\left( \dfrac{{{\text{A}}^{-}}}{\text{HA}} \right)$ is a Henderson - Hasselbalch equation.
Note: The main application of this equation is to determine the pH of a solution which consists of an acid and its conjugate base. In the equation, $\text{p}{{\text{K}}_{\text{a }}}$is the acid dissociation constant.
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