Question

potassium hydroxide KOH (base). The dissociation of KOH is shown below. If the Kb of the KOH base was $9.5 \times 10^{-3}$, calculate the equilibrium concentration of $[KOH]$, $[H_3O^+]$, $pOH$, $[OH^-]$ and $K_a$ of the liquid. Use ICE Table & The Approximation Rule if Possible $KOH(aq) + H_2O(l) \rightarrow OH^-(aq) + K^+(aq)$

          potassium hydroxide KOH (base). The dissociation of KOH is shown below. If the Kb of the KOH
base was $9.5 \times 10^{-3}$, calculate the equilibrium concentration of $[KOH]$, $[H_3O^+]$, $pOH$, $[OH^-]$ and
$K_a$ of the liquid. Use ICE Table & The Approximation Rule if Possible
$KOH(aq) + H_2O(l) \rightarrow OH^-(aq) + K^+(aq)$

potassium hydroxide KOH (base). The dissociation of KOH is shown below. If the Kb of the KOH
base was 9.5 × 10^-3, calculate the equilibrium concentration of [KOH], [H3O^+], pOH, [OH^-] and
Ka of the liquid. Use ICE Table The Approximation Rule if Possible
KOH(aq) + H2O(l) → OH^-(aq) + K^+(aq)

Added by Nicol-S B.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Chareen Guzman

08/25/2023

Step 1

Step 1: Write the equilibrium expression for the dissociation of KOH: KOH(aq) + H2O(l) ⇌ OH-(aq) + K+(aq) Kb = [OH-][K+]/[KOH] Show more…

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potassium hydroxide KOH (base). The dissociation of KOH is shown below. If the Kb of the KOH base was 9.5x10^(-3), calculate the equilibrium concentration of [KOH],[H_(3)O^(+)],pOH,[OH^(-)]and Ka of the liquid. Use ICE Table & The Approximation Rule if Possible KOH(aq)+H_(2)O(l)->OH_((aq))^(-)+K_((aq))^(+) potassium hydroxide KOH (base). The dissociation of KOH is shown below. If the Kb of the KOH base was 9.5 x10-3,calculate the equilibrium concentration of [KOH], [H30*3, pOH, [OH] and Ka of the liquid. Use ICE Table & The Approximation Rule if Possible KOH(aq)+H,O(l) > OH"(aq)+K*(aq)