Question

What is the `"ppm"` concentration of hydroxide in a `1.0xx10^-4*mol*L^-1` solution of calcium hydroxide?

Answer 1

Well, `1*"ppm"` specifies a concentration of `1*mg*L^-1`....

Explanation:

And so we address the concentration of `HO^-` in the solution....

And given a concentration of `1.0xx10^-4*mol*L^-1` with respect to `Ca(OH)_2`...I make this....

`1xx10^-4*mol*L^-1xx2xx17*g*mol^-1=3.4xx10^-3*g*L^-1..........................`

WITH RESPECT TO HYDROXIDE ION (which is why I introduced the factor of 2.....)

And so we got `3.4*"ppm"` with respect to hydroxide anion....

Answer 2

`["OH"^"-"] = "3.4 ppm"`

Explanation:

`["OH"^"-"] = (1.0 × 10^"-4" color(red)(cancel(color(black)("mol Ca(OH)"_2))))/(1 "L") × (2 color(red)(cancel(color(black)("mol OH"^"-"))))/(1 color(red)(cancel(color(black)("mol Ca(OH)"_2)))) × (17.01 "g OH"^"-")/(1 color(red)(cancel(color(black)("mol OH"^"-"))))`

`= 3.4 × 10^"-3"color(white)(l) "g/L" = "3.4 mg/L"`

`"1 ppm" = (1 "mg solute")/(10^6color(white)(l) "mg solution")`

∴ `["OH"^"-"] = "3.4 mg OH"^"-"/"1 L solution" = "3.4 mg OH"^"-"/"1000 g solution" = ("3.4 mg OH"^"-")/(10^6color(white)(l) "mg solution") = "3.4 ppm"`