What volume, in mL, of 0.100 M NaOH is required to neutralize 35.0 mL of 0.102 M hydroiodic acid HI?

35.7 mL NaOH

Explanation:

M1V1 = M2V2

M1 = 0.100 M NaOH

V1 = ?

M2 = 0.102 M HI

V2 = 35.0 mL

Solve for V1 --> V1 = M2V2/M1

V1 = (0.102 M)(35.0 mL) / (0.100 M) = 35.7 mL NaOH

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."