Question

How many grams of `SrO` should be dissolved in sufficient water to make 2.00 L of a solution with a pH= 11.00?

Answer 1

You should use 0.0641 g of `"SrO"`.

Explanation:

Concentration of `["OH"]^"-"`

`"pH" = 11.00`

∴ `"pOH" = "14 00 - 11.00" = 3.00`

`["OH"^"-"] = 10^"-pOH"color(white)(l) "mol/L" = 10^"-3.00"color(white)(l) "mol/L" = 1.00 × 10^"-3"color(white)(l)"mol/L"`

Mass of `"SrO"`

`"Moles of OH"^"-" = 2.00 color(red)(cancel(color(black)("L"))) × (1.00 × 10^"-3"color(white)(l) "mol OH"^"-")/(1 color(red)(cancel(color(black)("L")))) = 2.00 × 10^"-3"color(white)(l) "mol OH"^"-"`

The net ionic equation for the reaction is

`"O"^"2-" + "H"_2"O" → "2OH"^"-"`

`"Moles of O"^"2-" = 2.00 × 10^"-3" color(red)(cancel(color(black)("mol OH"^"-"))) × ("1 mol O"_2^"-")/(2 color(red)(cancel(color(black)("mol OH"^"-")))) = 1.00 × 10^"-3"color(white)(l) "mol O"^"2-"`

`"Moles of SrO" = 1.00 × 10^"-3" color(red)(cancel(color(black)("mol O"^"2-"))) × "1 mol SrO"/(1 color(red)(cancel(color(black)("mol O"^"2-")))) = 1.00 × 10^"-3"color(white)(l) "mol SrO"`

`"Mass of SrO" = 1.00 × 10^"-3" color(red)(cancel(color(black)("mol SrO"))) × "64.06 g SrO"/(1 color(red)(cancel(color(black)("mol SrO")))) = "0.0641 g SrO"`