If 15% of a O.1 M solution of a weak acid dissociates into H+ and A- ions, what is the concentration of H3O+? (HA + H2O --> H3O+ + A-) ?

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If 15% of a O.1 M solution of a weak acid dissociates into H+ and A- ions, what is the concentration of H3O+? (HA + H2O --> H3O+ + A-) ?

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Answer 1 · 2018-05-03T15:26:52

Would it not be $0.015 \cdot m o l \cdot L^{- 1}$ $0.015*mol*L^-1$ with respect to $H_{3} O^{+}$ $H_3O^+$ ?

Explanation:

And so we should address the equilibrium...

$H A (a q) + H_{2} O (l) ⇌ H_{3} O^{+} + A^{-}$ $HA(aq) +H_2O(l)rightleftharpoonsH_3O^+ + A^(-)$

Initially, $[H A] = 0.1 \cdot m o l \cdot L^{- 1}$ $[HA]=0.1*mol*L^-1$ ...finally... $[H_{3} O^{+}] \equiv [A^{-}] \equiv 0.015 \cdot m o l \cdot L^{- 1}$ $[H_3O^+]-=[A^-]-=0.015*mol*L^-1$ ...and $[H A] \equiv 0.1 - 0.015,$ $[HA]-=0.1-0.015,$ and we could work out $K_{a}$ $K_a$ from these data...

$K_{a} = \frac{[H_{3} O^{+}] [A^{-}]}{[H A]} = \frac{0.015 \times 0.015}{0.1 - 0.015} = 2.65 \times 10^{- 3}$ $K_a=([H_3O^+][A^-])/([HA])=(0.015xx0.015)/(0.1-0.015)=2.65xx10^-3$ ...

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If 15% of a O.1 M solution of a weak acid dissociates into H+ and A- ions, what is the concentration of H3O+? (HA + H2O --> H3O+ + A-) ?

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