How many moles of Ba(OH)2 are present in 115 mL of 0.400 M Ba(OH)2

A chemist needs the concentration of HNO3.
She puts 755 mL of the acid in a flask. She then adds 0.400 M Ba(OH)2 to the flask. She notes that 115 mL of Ba(OH)2 was needed to reach the equivalence point.
Solution map
In this titration, the concentration of base is known and can be used to calculate the unknown acid concentration:

concentration of base -> moles of base -> moles of acid -> concentration of acid

Ba(OH)2 + 2 HNO3 → Ba(NO3)2 + 2 H2O

(115 mL) x (0.400 M Ba(OH)2) x (2/1) / (755 mL) = 0.122 M HNO3