pH of acids, bases and salts

Exercise 9

   

Calculate the $pH$ of a 0.315 $\frac{g}{L}$  potassium hydroxide solution

$OH^-$ (treated as) strong base The number of moles of $OH^-$ is equal to the number of moles of potassium hydroxide ! Number of moles potassium hydroxide in $1\;L$: $n$ $=$ $\frac{m}{M}$ = (with $m$ $ =$ 0.315 $\; g$ and $M$ = 68.05) 0.0046$ \;mol$ $c_{OH^-}$ $=$ $\frac{n}{V}$ = 0.0046 $\frac{mol}{L}$ $pH$ $=$ $14$ $+$ $log\;c_{OH^-}$ = $14$ $+$ $log \;$0.0046 = 11.67