Question

Question #9e3a5

Answer 1

Working from the following metathesis reaction ...

`2HNO_3(aq) + Ca(OH)_2(aq)` => `Ca(NO_3)_2 + 2H_2O(l)`

The reaction ratio to neutralization is 2 moles `HNO_3` to 1 mole `Ca(OH)_2`

From definition of
`"Molarity"(M) = ("moles"."solute")/("Liters"."solution")`

=> `"moles"."solute"= "Molarity(M)"xx"Volume(L)"`

Therefore, one can use the relationship

1 mole `HNO_3` = 2 moles `Ca(OH)_2`

(Molarity x Volume)`HNO_3` = 2(Molarity x Volume)`Ca(OH)_2`

`(0.177M)HNO_3(aq) xx(28.2ml)HNO_3 "solution")` = `2[Ca(OH)_2(aq)] xx(17.5ml)Ca(OH)_2 "solution")`

Simplifying notations for illustration

(0.177M)(28.2ml) = `2[Ca(OH)_2(aq)]`(17.5ml)

`[Ca(OH)_2]` = `((2(0.177M)(28.2ml))/(17.5ml))`

= `0.570M Ca(OH)_2(aq) "solution"`

Note: In this problem case, the volume values may remain in milliliter dimensions as both sides of the (MV)acid = (MV)base are in ml's.