How do you calculate ${{\text{K}}_{\text{a}}}$?
Answer
600.6k+ views
Hint: For the calculation of the value of ${{\text{K}}_{\text{a}}}$, we have to know about the molar concentration of each species present on the left hand side as well as on the right hand side of the chemical reaction at the equilibrium state.
Complete answer:
Here ${{\text{K}}_{\text{a}}}$ is also known by the name of acid dissociation constant at the equilibrium condition which is equal to the ratio of the product of molar concentrations of the conjugate base of acid and hydronium ion to the molar concentration of the given weak acid.
Let us consider a weak acid (${\text{HA}}$) dissociates into the hydronium ion (${{\text{H}}^{\text{ + }}}$) and conjugate base (${{\text{A}}^{\text{ - }}}$) of the given acid at equilibrium condition and this reaction is given below:
${\text{HA}} \rightleftharpoons {{\text{H}}^{\text{ + }}}{\text{ + }}{{\text{A}}^{\text{ - }}}$
And for the above given chemical reaction, equation for the calculation of acid dissociation constant (${{\text{K}}_{\text{a}}}$) is written as:
${{\text{K}}_{\text{a}}}{\text{ = }}\dfrac{{\left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{{\text{A}}^{\text{ - }}}} \right]}}{{\left[ {{\text{HA}}} \right]}}$.
Additional information:
Acid dissociation constant (${{\text{K}}_{\text{a}}}$) is a unit less quantity because in the numerical and denominator of the above equation same unit of concentration is written which get cancelled from each other, that’s why it doesn’t have any unit.
Note:
Sometimes dissociation constant is also given by the name of equilibrium constant, so never get confused that equilibrium constant is a different quantity. Dissociation constant and equilibrium constant is the same quantity because it is calculated at the equilibrium state of the chemical reaction.
Complete answer:
Here ${{\text{K}}_{\text{a}}}$ is also known by the name of acid dissociation constant at the equilibrium condition which is equal to the ratio of the product of molar concentrations of the conjugate base of acid and hydronium ion to the molar concentration of the given weak acid.
Let us consider a weak acid (${\text{HA}}$) dissociates into the hydronium ion (${{\text{H}}^{\text{ + }}}$) and conjugate base (${{\text{A}}^{\text{ - }}}$) of the given acid at equilibrium condition and this reaction is given below:
${\text{HA}} \rightleftharpoons {{\text{H}}^{\text{ + }}}{\text{ + }}{{\text{A}}^{\text{ - }}}$
And for the above given chemical reaction, equation for the calculation of acid dissociation constant (${{\text{K}}_{\text{a}}}$) is written as:
${{\text{K}}_{\text{a}}}{\text{ = }}\dfrac{{\left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{{\text{A}}^{\text{ - }}}} \right]}}{{\left[ {{\text{HA}}} \right]}}$.
Additional information:
Acid dissociation constant (${{\text{K}}_{\text{a}}}$) is a unit less quantity because in the numerical and denominator of the above equation same unit of concentration is written which get cancelled from each other, that’s why it doesn’t have any unit.
Note:
Sometimes dissociation constant is also given by the name of equilibrium constant, so never get confused that equilibrium constant is a different quantity. Dissociation constant and equilibrium constant is the same quantity because it is calculated at the equilibrium state of the chemical reaction.
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