Question

What volume of 0.0557 M Sr(OH)2 solution is required to neutralize 36.4 mL of a 0.0750 M of HNO3 solution. a 36.4 mL b 49.0 mL c 98.0 mL d 24.5 mL

Answer 1

The volume of 0.0557 M
`SrOH_(2)` solution required to neutralize 36.4 mL of 0.0750 M
`HNO_(3)` solution is 73.8 mL.

To determine the volume of 0.0557 M
`Sr(OH)_(2)` solution needed to neutralize 36.4 mL of 0.0750 M
`HNO_(3)` solution, we can use the equation:


`HNO_(3) + Sr(OH)_(2) -- > H_(2) O + Sr(NO_(3) )_(2)`

From the equation, we can see that the molar ratio between
`HNO_(3)` and
`Sr(OH)_(2)` is 1:2.

Therefore, to neutralize the
`HNO_(3)`, we need twice as much
`Sr(OH)_(2)`. Convert the volume of
`HNO_(3)` to moles using the molarity, and then use the mole ratio to calculate the volume of
`Sr(OH)_(3)` solution needed.

Since we need twice as much
`Sr(OH)_(2)`, the volume required would be 2 times the volume of
`HNO_(3)`, which is 73.8 mL.