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?

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

Where is the graph ?

Since there is the same amount of agn03 in both silver chloride and silver nitrate they produce the same amount of moles. therefore silver chloride produces the same amount of silver nitrate, so your answer is 15.0 moles of silver nitrate. i hope this !