Calculate the $pH$ of a $c_o$ $=$ $3.5 \cdot 10^{-5}\;M$ solution of the strong base $Sr(OH)_2$: $Sr(OH)_2$ $\longrightarrow$ $Sr^{2+}$ $+$ $2 OH^-$ What equation do we need to calculate the $pH$ ?
a) $pH$ $=$ $[H_3O^+][OH^-]$ b) $pH$ $=$ $- log [H_3O^+]$ c) $pH$ $=$ $\frac{[OH^-]}{ [H_3O^+] }$ d) $pH$ $=$ $- log [OH^-]$
The answer is b) $pH$ $=$ $- log [H_3O^+]$ $=$ $9.85$