The titration reaction is written as follows:
Ba(OH)2 + 2HNO3 = Ba(NO3)2 + 2H2O
The calculations are as done as follows:
0.100 mol HNO3 / L (.03445 L solution ) = 0.003445 mol HNO3
0.003445 mol HNO3 ( 1 mol Ba(OH)2 / 2 mol HNO3 ) = 0.001723 mol Ba(OH)2
0.001723 mol / 0.025 L = 0.07 M Ba(OH)2
