## Chemistry and Chemical Reactivity (9th Edition)

$29.10\ mL$
Number of moles of $OH^-$: $200.\ mg\div78.00\ mg/mmol\times3+200.\ mg\div58.32\ mg/mmol\times2=14.55\ mmol$ Same number of moles of acid to complete neutralization: $14.55\ mmol\div0.500\ M=29.10\ mL$