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?
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.
To find the molarity (M) of the original calcium hydroxide solution, we will use the concept of stoichiometry and the neutralization reaction between calcium hydroxide (Ca(OH)2) and hydroiodic acid (HI).
First, we need to determine the balanced chemical equation for the reaction:
Ca(OH)2 + 2HI -> CaI2 + 2H2O
From the balanced equation, we can see that 1 mole of calcium hydroxide reacts with 2 moles of hydroiodic acid. Therefore, their molar ratio is 1:2.
Given that 31.7 mL of 1.95 M hydroiodic acid is added, we can calculate the number of moles of hydroiodic acid:
moles of HI = volume (L) x concentration (M)
= 31.7 mL x 0.001 L/mL x 1.95 M
= 0.061995 mol
Since the molar ratio of Ca(OH)2 to HI is 1:2, the number of moles of calcium hydroxide present is half of the number of moles of hydroiodic acid:
moles of Ca(OH)2 = 0.061995 mol ÷ 2
= 0.0309975 mol
Now, we can determine the molarity of the original calcium hydroxide solution using the given volume:
Molarity (M) = moles ÷ volume (L)
= 0.0309975 mol ÷ (42.3 mL x 0.001 L/mL)
= 0.73184 M
Therefore, the molarity of the original calcium hydroxide solution is approximately 0.73184 M.