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What is the equation for the base ionization constant of RbOH?

1 Answer

Jul 3, 2018

$K_{b} = \frac{[R b^{+}] [O H^{-}]}{R b O H}$ $K_b = ([Rb^+][OH^-])/[RbOH]$

Explanation:

The equation for ionization of rubidium hydroxide, RbOH can be written as

$R b O H (a q) \to R b^{+} (a q) + O H^{-}$ $RbOH(aq) -> Rb^+(aq) + OH^-$ $(a q)$ $(aq)$

From the equation, the base ionization constant, $K_{b}$ $K_b$ , can be written as

$K_{b} = \frac{[R b^{+}] [O H^{-}]}{R b O H}$ $K_b = ([Rb^+][OH^-])/[RbOH]$

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