Calculate number of ions in 5.25g of aluminium sulfate?
$Al_{2}(SO_{4})_{3} \to 5$ ions
$n = \frac{m}{M}, \therefore n(Al_{2}(SO_{4}){3}) = \frac{5.25}{2 \times 26.98 + 3(32.06 + 4 \times 16)} = 0.0153446mol$ 2 mols of $Al$ and 3 mols of $SO{4}$ per mol of $Al_{2}(SO_{4}){3}$, $\therefore n(ions) = 5 \times n(Al{2}(SO_{4})_{3}) = 0.0767mol (3SF)$
$25.0 mL$ of $0.73 molL^{-1}$ of $HCl$ added to $30.0mL$ of $0.999molL^{-1}$ of $NaOH$. Calculate pH of the resulting solution
