The hydrogen ion concentration in a solution of weak acid of dissociation constant Ka and concentration C is nearly equal to:

Question

The hydrogen ion concentration in a solution of weak acid of dissociation constant Ka and concentration C is nearly equal to: (A)  Ka.C  (B)  C/Ka  (C)  √Ka.C  (D)  √Ka/C

Tardigrade Admin · Accepted Answer

HA leftharpoons H ++A- Ka=(C2 x2/C(a-x)) (C2 x2/C) (For weak acid .1-x ≅ 1) C ⋅ Ka=C2 x2=C x=√Ka ⋅ C ∴ [ H +]= C x=√Ka ⋅ C

Q. The hydrogen ion concentration in a solution of weak acid of dissociation constant $K_{a}$ and concentration $C$ is nearly equal to:

$K_{a} . C$

$C / K_{a}$

$K_{a} . C$

$K_{a} / C$

$H A ⇌ H^{+} + A^{-}$
$K_{a} = \frac{C ^{2} x ^{2}}{C ( a - x )}$
$\frac{C _{2} x ^{2}}{C}$
(For weak acid $1 - x ≅ 1)$
$C \cdot K_{a} = C^{2} x^{2} = C x = K_{a} \cdot C$
$∴ [H^{+}] = C x = K_{a} \cdot C$

Q. The hydrogen ion concentration in a solution of weak acid of dissociation constant Ka​ and concentration C is nearly equal to:

Ka​.C

C/Ka​

Ka​.C​

Ka​/C​

HA⇌H++A− Ka​=C(a−x)C2x2​ CC2​x2​ (For weak acid 1−x≅1) C⋅Ka​=C2x2=Cx=Ka​⋅C​ ∴[H+]=Cx=Ka​⋅C​

Q. The hydrogen ion concentration in a solution of weak acid of dissociation constant $K_{a}$ and concentration $C$ is nearly equal to:

$K_{a} . C$

$C / K_{a}$

$K_{a} . C$

$K_{a} / C$

$H A ⇌ H^{+} + A^{-}$
$K_{a} = \frac{C ^{2} x ^{2}}{C ( a - x )}$
$\frac{C _{2} x ^{2}}{C}$
(For weak acid $1 - x ≅ 1)$
$C \cdot K_{a} = C^{2} x^{2} = C x = K_{a} \cdot C$
$∴ [H^{+}] = C x = K_{a} \cdot C$