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# pH of strong bases

Neglecting the contribution of hydroxide ions from the autoprotolysis of water (approximation), it can be said that each hydroxide ion present in the solution comes from a strong base molecule, thus the number of moles of base (per liter) prior to the separation is equal to the number of moles (per liter) hydroxide ion after dissociation: $c_{B}$ $=$ $[OH^-]$ therefore:

Strong bases: $pOH$ $=$ $-log\;c_{B}$ $pH$ $=$ $14$ $+$ $log\;c_{B}$

Examples: The $pH$ of a $0.2\; M$ solution of sodium amide $NaNH_2$ (which forms by first dissociation $NH_2^-$) is: $pH$ $=$ $14$ $+$ $log0.2$ $=$ $13.3$ The $pH$ of a $0.1\; M$ sodium hydroxide solution $NaOH$ (which forms by dissociation $OH^-$) is: $pH$ $=$ $14$ $+$ $log0.1$ $=$ $13.0$

Special cases: $O^{2-}$ and $Me(OH)_2$ produce 2 hydroxides : $2c_{B}$ $=$ $[OH^-]$, therefore: The $pH$ of a $0.05\; M$ sodium oxide $Na_2O$ solution (2 hydroxides formed by dissociation of $O^{2-}$) is: $pH$ $=$ $14$ $+$ $log(2\cdot0.5)$ $=$ $13.0$ The $pH$ of a $0.1 \;M$ solution of calcium hydroxide $Ca(OH)_2$ solution (2 hydroxides formed by dissociation) is: $pH$ $=$ $14$ $+$ $log(2\cdot0.1)$ $=$ $13.3$