$pH$ of strong bases

Tutorial 7

        

Calculation of the pH of a strong base Case of soluble mineral hydroxides: From the initial molarity of the hydroxide, $ [OH^-] $ is calculated after the complete dissociation $pOH=-log\;[OH ^-]$ $pH=14-pOH$ Case of other strong bases: Since the protolysis is complete, each strong base mole $ B $ has given one mole of hydronium ( except $O^{2-}$ which gives two!) therefore $ [OH^-] = c_{B} $ Where $ c_{B} = \frac{n_ {B}}{V} $ is the initial molarity of the strong base. $pOH=-log\;c_B$ $pH=14-pOH$

Calculate the pH of a solution $ c_o $ $ = $ $ 3.5 \cdot 10^{-5} M $ of the strong base, $ Sr(OH)_2 $: $ Sr(OH)_2 $ $ \longrightarrow $ $Sr^{2+} $ $ + $ $ 2 OH^- $ Calculate $ [H_3O^+] $ as in the previous exercise, then answer: What equation do we use to calculate the $pH$?

a) $pH$ $=$ $[H_3O^+][ OH^-]$ b) $pH$ $=$ $- log [H_3O^+]$ c) $pH$ $=$ $\frac{[ OH^-]}{ [H_3O^+] }$ d) $pH$ $=$ $- log [ OH^-]$

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