The first and second ionization constants of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the first ionization step of the weak dibasic acid: $H_{2}A \rightleftharpoons H^{+} + HA^{-}$.The equilibrium constant expression is $K_{a1}

Vedclass · Accepted Answer

$8.1 	imes 10^{-8} 	ext{ M}$

Vedclass · Answer

(C) For the first ionization step of the weak dibasic acid: $H_{2}A ightleftharpoons H^{+} + HA^{-}$.The equilibrium constant expression is $K_{a1} = \frac{[H^{+}][HA^{-}]}{[H_{2}A]}$.Given that $0.1 	ext{ mol}$ of $H_{2}A$ is dissolved in $1 	ext{ L}$ of solution, the initial concentration $[H_{2}A] = 0.1 	ext{ M}$.The solution also contains $0.1 	ext{ M}$ $HCl$, which is a strong acid and dissociates completely, providing $[H^{+}] = 0.1 	ext{ M}$.Since $K_{a1} = 8.1 	imes 10^{-8}$ is very small, the dissociation of $H_{2}A$ is negligible. Therefore, we can assume $[H_{2}A] \approx 0.1 	ext{ M}$ and $[H^{+}] \approx 0.1 	ext{ M}$ at equilibrium.Substituting these values into the expression: $8.1 	imes 10^{-8} = \frac{(0.1)[HA^{-}]}{0.1}$.Solving for $[HA^{-}]$, we get $[HA^{-}] = 8.1 	imes 10^{-8} 	ext{ M}$.

The first and second ionization constants of a weak dibasic acid $H_{2}A$ are $8.1 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. $0.1 \text{ mol}$ of $H_{2}A$ was dissolved in $1 \text{ L}$ of $0.1 \text{ M}$ $HCl$ solution. The concentration of $HA^{-}$ in the resultant solution is:

Similar Questions

The degree of dissociation of a $0.05 \ M$ solution of $NH_3$ having $pH = 11$ at $25 \ ^oC$ is ...............

What is the ratio of the number of $H^{+}$ ions to the number of $H_2O$ molecules in $1 \ L$ of water?

Given,$pH$ of a solution $A$ is $3$ and it is mixed with another solution $B$ having $pH$ $2$. If both are mixed in equal volume,then the resultant $pH$ of the solution will be

At $25^{\circ} C$, the molar conductance of $0.007 \ M$ hydrofluoric acid is $150 \ \text{mho} \ cm^{2} \ mol^{-1}$ and its $\Lambda_{m}^{\circ} = 500 \ \text{mho} \ cm^{2} \ mol^{-1}$. The value of the dissociation constant of the acid at the given concentration at $25^{\circ} C$ is

Assign $A, B, C, D$ from the given type of reaction.
$Cu(OH)_2 \downarrow + 4NH_3(soln.) \longrightarrow [Cu(NH_3)_4]^{2+} + 2OH^{-}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module