How do you determine the \[{{\text{K}}_{\text{a}}}\] of weak acid and \[{{\text{K}}_{\text{b}}}\] of weak base?
Answer
624k+ views
Hint: To answer this question we should know what is \[{{\text{K}}_{\text{a}}}\] ,\[{{\text{K}}_{\text{b}}}\] .These are used to determine the strength of acids and bases. \[{{\text{K}}_{\text{a}}}\] is ionization constant for weak acids and \[{{\text{K}}_{\text{b}}}\] is ionization constant for weak base. So depending on these acids and bases can be weak and strong. If an acid or base is completely ionized in aqueous solution it is a strong acid or strong base. But if it is slightly ionized then it is a weak acid or weak base.
Complete step by step answer:
As we know that the reaction of weak acid with water produces \[{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}\] or \[{{\text{H}}^{\text{ + }}}\]
To determine acid ionization constant \[{{\text{K}}_{\text{a}}}\] , following is the equation for ionization of weak acid $\left[ {HA} \right]$
in water and [\[{{\text{A}}^{\text{ - }}}\]] is the conjugate base.
\[{\text{HA(aq) + }}{{\text{H}}_{\text{2}}}{\text{O(l)}} \rightleftarrows {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{(aq) + }}{{\text{A}}^{\text{ - }}}{\text{(aq)}}\]
The equilibrium constant for the equation can be written as,
\[{\text{K = }}\dfrac{{{\text{[}}{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{][}}{{\text{A}}^{\text{ - }}}{\text{]}}}}{{{\text{[}}{{\text{H}}_{\text{2}}}{\text{O][HA]}}}}\]
Since concentration of water is constant so \[{\text{K}}\] can be written as \[{\text{K[}}{{\text{H}}_{\text{2}}}{\text{O]}}\] and new constant form is \[{{\text{K}}_{\text{a}}}\]and also known as acid dissociation constant.
\[{{\text{K}}_{\text{a}}}{\text{ = K[}}{{\text{H}}_{\text{2}}}{\text{O] = }}\dfrac{{{\text{[}}{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{][}}{{\text{A}}^{\text{ - }}}{\text{]}}}}{{{\text{[HA]}}}}\]
We can write \[{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}\] as \[{{\text{H}}^{\text{ + }}}\] but no \[{{\text{H}}^{\text{ + }}}\] ion exist in aqueous medium so we use \[{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}\].
Now weak bases on reaction with water produces \[{\text{O}}{{\text{H}}^{\text{ - }}}\] (hydroxide) ions.
To determine base ionization constant \[{{\text{K}}_{\text{b}}}\] ,following is the equation for ionization of weak base $\left[ B \right]$ in water and [\[{\text{B}}{{\text{H}}^{\text{ + }}}\]] is conjugate acid.
\[{\text{B(aq) + }}{{\text{H}}_{\text{2}}}{\text{O(l)}} \rightleftarrows {\text{B}}{{\text{H}}^{\text{ + }}}{\text{(aq) + O}}{{\text{H}}^{\text{ - }}}{\text{(aq)}}\]
The equilibrium constant for the equation can be written as,
\[{{\text{K}}_b}{\text{ = K[}}{{\text{H}}_{\text{2}}}{\text{O] = }}\dfrac{{{\text{[B}}{{\text{H}}^{\text{ + }}}{\text{][O}}{{\text{H}}^{\text{ - }}}{\text{]}}}}{{{\text{[B]}}}}\]
If we know the value of \[{{\text{K}}_{\text{a}}}\] and \[{{\text{K}}_{\text{b}}}\] we can calculate \[{\text{p}}{{\text{K}}_{\text{a}}}\] and \[{\text{p}}{{\text{K}}_{\text{b}}}\]
Note:
We should be noted that acid- base ionization constants are measured in terms of \[{{\text{H}}^{\text{ + }}}\] and \[{\text{O}}{{\text{H}}^{\text{ - }}}\] and therefore they don’t have any unit. Example of strong acid is \[{\text{HCl}}\] and \[{\text{C}}{{\text{l}}^{\text{ - }}}\] (chloride ion) is its weak conjugate base. Similarly \[{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\] is weak acid and its strong base is \[{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^{\text{ - }}}\] (acetate ion).It should be noted that the larger the \[{{\text{K}}_{\text{a}}}\] ,the stronger is the acid(\[{\text{p}}{{\text{K}}_{\text{a}}}\]). Similarly the larger the \[{{\text{K}}_{\text{b}}}\] , the stronger is the base (\[{\text{p}}{{\text{K}}_{\text{b}}}\]) .
Complete step by step answer:
As we know that the reaction of weak acid with water produces \[{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}\] or \[{{\text{H}}^{\text{ + }}}\]
To determine acid ionization constant \[{{\text{K}}_{\text{a}}}\] , following is the equation for ionization of weak acid $\left[ {HA} \right]$
in water and [\[{{\text{A}}^{\text{ - }}}\]] is the conjugate base.
\[{\text{HA(aq) + }}{{\text{H}}_{\text{2}}}{\text{O(l)}} \rightleftarrows {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{(aq) + }}{{\text{A}}^{\text{ - }}}{\text{(aq)}}\]
The equilibrium constant for the equation can be written as,
\[{\text{K = }}\dfrac{{{\text{[}}{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{][}}{{\text{A}}^{\text{ - }}}{\text{]}}}}{{{\text{[}}{{\text{H}}_{\text{2}}}{\text{O][HA]}}}}\]
Since concentration of water is constant so \[{\text{K}}\] can be written as \[{\text{K[}}{{\text{H}}_{\text{2}}}{\text{O]}}\] and new constant form is \[{{\text{K}}_{\text{a}}}\]and also known as acid dissociation constant.
\[{{\text{K}}_{\text{a}}}{\text{ = K[}}{{\text{H}}_{\text{2}}}{\text{O] = }}\dfrac{{{\text{[}}{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{][}}{{\text{A}}^{\text{ - }}}{\text{]}}}}{{{\text{[HA]}}}}\]
We can write \[{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}\] as \[{{\text{H}}^{\text{ + }}}\] but no \[{{\text{H}}^{\text{ + }}}\] ion exist in aqueous medium so we use \[{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}\].
Now weak bases on reaction with water produces \[{\text{O}}{{\text{H}}^{\text{ - }}}\] (hydroxide) ions.
To determine base ionization constant \[{{\text{K}}_{\text{b}}}\] ,following is the equation for ionization of weak base $\left[ B \right]$ in water and [\[{\text{B}}{{\text{H}}^{\text{ + }}}\]] is conjugate acid.
\[{\text{B(aq) + }}{{\text{H}}_{\text{2}}}{\text{O(l)}} \rightleftarrows {\text{B}}{{\text{H}}^{\text{ + }}}{\text{(aq) + O}}{{\text{H}}^{\text{ - }}}{\text{(aq)}}\]
The equilibrium constant for the equation can be written as,
\[{{\text{K}}_b}{\text{ = K[}}{{\text{H}}_{\text{2}}}{\text{O] = }}\dfrac{{{\text{[B}}{{\text{H}}^{\text{ + }}}{\text{][O}}{{\text{H}}^{\text{ - }}}{\text{]}}}}{{{\text{[B]}}}}\]
If we know the value of \[{{\text{K}}_{\text{a}}}\] and \[{{\text{K}}_{\text{b}}}\] we can calculate \[{\text{p}}{{\text{K}}_{\text{a}}}\] and \[{\text{p}}{{\text{K}}_{\text{b}}}\]
Note:
We should be noted that acid- base ionization constants are measured in terms of \[{{\text{H}}^{\text{ + }}}\] and \[{\text{O}}{{\text{H}}^{\text{ - }}}\] and therefore they don’t have any unit. Example of strong acid is \[{\text{HCl}}\] and \[{\text{C}}{{\text{l}}^{\text{ - }}}\] (chloride ion) is its weak conjugate base. Similarly \[{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\] is weak acid and its strong base is \[{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^{\text{ - }}}\] (acetate ion).It should be noted that the larger the \[{{\text{K}}_{\text{a}}}\] ,the stronger is the acid(\[{\text{p}}{{\text{K}}_{\text{a}}}\]). Similarly the larger the \[{{\text{K}}_{\text{b}}}\] , the stronger is the base (\[{\text{p}}{{\text{K}}_{\text{b}}}\]) .
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

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

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

