millimols HI used initially = mL x M = 31.7 x 1.95 = 61.8 but that was over titrated and showed an excess of HI. It was titrated with 17.3 mL of 1.00 M = 17.3 mmols Ba(OH)2 to determine how much of an excess was there.
.......................Ba(OH)2 + 2HI ==> BaI2 + 2H2O
I........................17.3..........61.8..........................
C.....................-17.3........-17.3*2..................
E.......................0..............27.2..............................
So 27.2 mmols of HI was the actual amount used in the titration of the Ca(OH)2.
.................. 2HI + Ca(OH)2 ==> CaI2 + 2H2O
I..................27.2........?....................................
C...............
E..................
If it took 27.2 mmols HI it must have taken 27.2/2 = 13.6 mmols Ca(OH)2.
Since M = millimols/mL then 13.6/42.3 mL = ?M Ca(OH)2.
Post your work if you get stuck.
31.7 mL of 1.95 M hydroiodic acid is added to 42.3 mL of calcium hydroxide, and the resulting solution is found to be acidic.
17.3 mL of 1.00 M barium hydroxide is required to reach neutrality.
What is the molarity of the original calcium hydroxide solution?
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1 answer