Here we have a $c_o$ $=$ $3.5 \cdot 10^{-5}\;M$ solution of the strong base $Sr(OH)_2$: $Sr(OH)_2$ $\longrightarrow$ $Sr^{2+}$ $+$ $2 OH^-$ Determine $[OH^-]$ !
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) $[OH^-]$ $=$ $2 c_0(Sr(OH)_2 )$ $=$ $7.0 \cdot 10^{-5} \;M$
Réponse : d) $[OH^-]$ $ =$ $ 2 c_0(Sr(OH)_2 )$ $= 7.0 \cdot 10^{-5}\;M$ because every mole of base dissociates to produce 2 moles of hydroxide ions.