Chemistry and Chemical Reactivity (9th Edition)

$0.33\ g$
Number of moles of the acid: $50.0\ mL \div 0.125\ M=6.25\ mmol$ From stoichiometry, number of moles of acid required: $0.00313\ mol$ Mass required: $0.00313\ mol\times 105.99\ g/mol=0.33\ g$