Question

What mass of Sr(OH)2 must be present in 750mL if the pH of this solution is 10.15?

Answer 1

first find the pOH from pH

pH + pOH = 14

pOH = 14 -pH = 14 - 10.15 = 3.85

from the pOH find the concentration of [OH-]

[OH-] = 10^-pOH = 10^-3.85 = 1.41 x 10^-5 M

moles of OH- = Molarity x volume in liters

moles of OH- = 1.41 x 10^-5 M x 0.75L = 1.06 x 10^-4 mol

lets write the dissociation equation of given base

Sr(OH)2 ---> Sr2+ + 2OH-

from balanced equation

1 mole of Sr(OH)2 is giving 2 moles of OH- accordingly

how many moles of Sr(OH)2 will give 1.06 x 10^-4 mol

= moles of Sr(OH)2 = 1.06 x 10^-4 mol / 2 = 5.3 x 10^-5 mol

Mass = Molaes x molar mass =

= 5.3 x 10^-5 mol x 121.6347 g/mol

= 0.00644 grams