The first and second ionization constants of | vedclass

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(A) For a diprotic acid $H_2X$, the ionization steps are: $H_2X ightleftharpoons H^{+} + HX^{-}$, $K_{a1} = 2.5 	imes 10^{-8}$ $HX^{-} ightleftha

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) For a diprotic acid $H_2X$, the ionization steps are: $H_2X ightleftharpoons H^{+} + HX^{-}$, $K_{a1} = 2.5 	imes 10^{-8}$ $HX^{-} ightleftharpoons H^{+} + X^{2-}$, $K_{a2} = 1.0 	imes 10^{-13}$ Since $K_{a1} \gg K_{a2}$, the concentration of $H^{+}$ is primarily determined by the first dissociation. For $0.1 \ M$ $H_2X$: $[H^{+}] \approx \sqrt{K_{a1} 	imes C} = \sqrt{2.5 	imes 10^{-8} 	imes 0.1} = \sqrt{2.5 	imes 10^{-9}} = 5 	imes 10^{-5} \ M$. The second dissociation constant is $K_{a2} = \frac{[H^{+}][X^{2-}]}{[HX^{-}]}$. Since $[H^{+}] \approx [HX^{-}]$, we have $[X^{2-}] = K_{a2} = 1.0 	imes 10^{-13} \ M$. Expressing this in the form $Y 	imes 10^{-15} \ M$: $1.0 	imes 10^{-13} = 100 	imes 10^{-15} \ M$. Thus, $Y = 100$.

Base	$K_{b}$
Dimethylamine,$(CH_{3})_{2}NH$	$5.4 \times 10^{-4}$
Triethylamine,$(C_{2}H_{5})_{3}N$	$6.45 \times 10^{-5}$
Ammonia,$NH_{3}$	$1.77 \times 10^{-5}$
Quinine	$1.10 \times 10^{-6}$
Pyridine,$C_{5}H_{5}N$	$1.77 \times 10^{-9}$
Aniline,$C_{6}H_{5}NH_{2}$	$4.27 \times 10^{-10}$
Urea,$CO(NH_{2})_{2}$	$1.3 \times 10^{-14}$

The first and second ionization constants of $H_2X$ are $2.5 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. The concentration of $X^{2-}$ in $0.1 \ M$ $H_2X$ solution is . . . . . . $\times 10^{-15} \ M$. The value of $Y$ is:

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