What volume of `0.00300 M Hcl` solution would just neutralize `30.0 mL` of `0.00100 M Ca(OH)_(2)` solution?

Options

A. `20.0 mL`
B. `30.0 mL`
C. `78.60 mL`
D. `25.22 mL`

Answer

1

Explanation

Write the balanced equation for the reaction to determine the reaction ratio.
`2Hcl + Ca(OH)_(2) rarr CaCl_(2) + 2H_(2)O`
`2 mmol 1 mmol 1 mmol 2mmol`
Then (1) convert milliliters of `Ca(OH)_(2)` solution to millimoles of `Ca(OH)_(2)` using molarity as a unit factor, `0.00100 mmol Ca(OH)//1.00 mL Ca(OH)_(2)` solution, (2) convert millimoles of `Ca(OH)_(2)` to millimoles of HCl using factor, `2 mmol HCl//1 mmol Ca(OH)_(2)`. (the reaction ratio from the balance equation), and (3) convert millimoles of HCl to milliliters of HCl solution using the unit factor, `1.00 mL HCl//0.00300 mmol HCl`, that is molarity inverted.

`30.0 mL Ca(OH)_(2)xx(0.00100 mmol Ca(OH)_(2))/(1.00 mL Ca(OH)_(2))xx(2 mmol HCl)/(1 mmol Ca(OH)_(2))xx(1.00 mL HCl)/(0.00300 mmol HCl) =20.0 mL HCl`
Note that the balance chemical equation allows us to construct either a mole ratio or a millimole ratio
`(2 mol HCl)/(1 mol Ca(OH)_(2))` or `(2 mmol HCl)/(1 mmol Ca(OH)_(2))`