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^- $ with: $ [OH^-] $ $ = $ $ 7.0 \cdot 10^{-5} M $ What is $ [H_3O^+] $?
a)$[H_3O^+]$ $=$ $[OH^-]$ b)$[H_3O^+]$ $=$ $2 [OH^-]$ c)$[H_3O^+]$ $=$ $\frac{[OH^-]}{K_e}$ d)$[H_3O^+]$ $=$ $\frac{K_e}{[OH^-]}$ ( with $K_e=10^{-14}\frac{mol^2}{L^2}$)
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Answer : d) $[H_3O^+]$ $ =$ $\frac{[K_e]}{[OH^-]}$
