2.
Given molarity of Ca(OH)2 = 1.50 M
Given volume of Ca(OH)2 = 45.0 mL = 0.045 L ( 1 mL = 0.001 L)
Hence, moles of Ca(OH)2 used = molarity of Ca(OH)2 x liter of Ca(OH)2
= 1.50 M x 0.045 L
= 0.0675 mol
Given, the volume HI titrated = 25.0 mL = 0.025 L
Let us say that the molarity of HI = z M
Hence, moles of HI titrated = molarity of HI x liter of HI
= z M x 0.025 L
= 0.025z mol
Ca(OH)2 + 2HI CaI2 + 2H2O
Now, from the balanced equation above, it is clear that 1 mol of Ca(OH)2 is required to neutralize 2 mol of HI .
Thus, equating the moles of Ca(OH)2 and HI to get
0.025z = 2 x 0.0675
or, z = 5.40 M
Hence, the molarity of HI = 5.40 M
titration: 2. If 45.0 mL of 1.50 M Ca(OH)2 are needed to neutralize 25.0 mL of...
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