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

Dec 11, 2016

You should use 0.0641 g of $\text{SrO}$.

#### Explanation:

Concentration of ["OH"]^"-"

$\text{pH} = 11.00$

$\text{pOH" = "14 00 - 11.00} = 3.00$

["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 $\text{SrO}$

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

The net ionic equation for the reaction is

$\text{O"^"2-" + "H"_2"O" → "2OH"^"-}$

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

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

$\text{Mass of SrO" = 1.00 × 10^"-3" color(red)(cancel(color(black)("mol SrO"))) × "64.06 g SrO"/(1 color(red)(cancel(color(black)("mol SrO")))) = "0.0641 g SrO}$