What volume of 0.378 M hydroiodic acid (HI) is required to neutralize 21.6 mL of 0.520...

What volume of 0.378 M hydroiodic acid (HI) is required to neutralize 21.6 mL of 0.520 M barium hydroxide Ba(OH)2?

Expert Solution

Ba(OH)2 + 2HI --> BaI2 + 2H2O

moles Ba(OH)2 present = 0.520 M x 21.6 ml = 11.232 mmol

moles HI needed = 2 x 11.232 mmol = 22.464 mmol

Volume HI solution needed for complete beutralization of Ba(OH)2 = 11.464 mmol/0.378 M

= 59.43 ml

queen_honey_blossom answered 2 years ago

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