Question

The hydrogen ion concentration in a weak acid of dissociation constant \[{K_a}\] ​and concentration $C$ is nearly equal to :
A. $\sqrt {\dfrac{{{K_a}}}{C}} $
B. $\dfrac{C}{{{K_a}}}$
C. ${K_a}C$
D. $\sqrt {{K_a}C} $

Answer 1

Hint: Using the value of \[{K_a}\] and the molar concentration of the weak acid, the hydrogen ion concentration may be computed. The stronger the acid and the higher the equilibrium \[{H^ + }\] concentration are, the larger the \[{K_a}\] .

Complete step-by-step answer:Suppose, \[C\] is the concentration of a given weak acid, $x$ is concentration of the ions obtained after dissociation of the acid and ${K_a}$ is the dissociation constant.
Now, consider a weak acid say \[HA\] that undergoes dissociation to give \[{H^ + }\] and \[{A^ - }\] .
 $HA\,\, \rightleftharpoons \,\,{H^ + }\, + \,{A^ - }$
Now the concentration of this weak acid is given as equal to \[C\] , that is, at time $t = 0$ , concentration of the given weak acid is \[C\] and obviously the concentration of \[{H^ + }\] and \[{A^ - }\] is 0 (before dissociation).
After some time during the reaction, the concentration of the given weak acid will be \[C - x\] and that of \[{H^ + }\] and \[{A^ - }\] is $x$ each.
As a result the dissociation constant becomes,
 ${K_a} = \dfrac{{x \times x}}{{C - x}}$
Since the acid given is weak, therefore, the value of $x$ is very less than 1. So we can ignore $x$ as compared to \[C\] in the denominator. The dissociation constant becomes
 ${K_a} = \dfrac{{{x^2}}}{C}$
Thus we get
 $x = \sqrt {{K_a}C} $ which is the required solution.

Option ‘D’ is correct

Note: An acid that partially separates into its ions in water or an aqueous solution is referred to as a weak acid. On the other hand, in water, a strong acid completely dissociates into its ions. While the conjugate acid of a weak base is also a weak acid, the conjugate base of a weak acid is also a weak base.