$pH$ of strong bases

Tutorial 9

    

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. $pOH=-log\;c_B$ $pH=14-pOH$

Let us calculate the pH of a solution of potassium methanolate $ CH_3OK \; \frac{g}{L}$: What is the chemical species that influences pH and why ?

For answers, use (possibly several times) the arrows ↑ Down! and ↓ Up! Complete please this question before moving on to the next one!