Here is a $ c_o = 3.5 \cdot 10^{-5} M $ solution of the strong base, $Sr(OH)_2$: $Sr(OH)_2$ $\longrightarrow$ $Sr^{2+}$ $+$ $2OH^-$
What is $[OH^-]$?
a) $[OH^-]$ $=$ $c_0( Sr(OH)_2)$ $=$ $3.5 \cdot 10^{-5} M$ b) $[OH^-]$ $=$ $[H_3O^+ ]$ $=$ $K_e^{\frac{1}{2}}$ $=$ $1.0 \cdot 10^{-7} M$ c) $[OH^-]$ $=$ $0,5\; c_0( Sr(OH)_2)$ $=$ $1.75 \cdot 10^{-5} M $ d) $[OH^-]$ $=$ $2 \;c_0(Sr(OH)_2)$ $=$ $7.0 \cdot 10^{-5} M$
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Réponse : d) $[OH^- ]$ $=$ $ 2 c_0(Sr(OH)_2)$ $=$ $7.0 \cdot 10^{-5} M$ since each mole of the base which dissociates forms 2 moles of the hydroxide ion.