How many milliliters of 0.120 m naoh are required to titrate 50.0 ml of 0.0998 m benzoic acid to the equivalence point? the ka of benzoic acid is 6.3 ⋅ 10-5?

Respuesta :

Answer: moles acid = 0.0500 L x 0.0998 = 0.00499 = moles NaOH V = 0.00499/ 0.120 = 0.0416 L >> 41.6 mL