What is the H+ concentration of a solutoin that is formed by combining 335 mL of 0.110 M HNO3 with 75 mL of 0.120 M Ba(OH)2 at 25 degrees celsius?
2HNO3 + Ba(OH)2 ==> Ba(NO3)2 + 2H2O
mols HNO3 = 0.110 x 0.335 = 0.03685
mols Ba(OH)2 = 0.00900
mols Ba(OH)2 is limiting reagent and will react 0.009 mols Ba(OH)2 to 0.018 (twice Ba(OH)2).
mols HNO3 in excess (unreacted) = 0.03685-0.018 = ?
M = ?mols/total Lsoln
Since HNO3 is a strong acid that will be the H^+.