Question

28 mL of 0.10 M HCl is added to 60 mL of 0.10 M Sr(OH)2. Using proper unit cancellation, determine the concentration of OH− in the resulting solution.

Answer 1

The moles of HCl added = Molarity of HCl X volume of HCl in litres

Moles of HCl = 0.1 X 28 / 1000 = 0.0028 moles

Moles of Sr(OH)2 present = Molarity X volume in litres

Moles of Sr(OH)2 = 0.1 x 60 / 1000 = 0.006 moles

The reaction equation will be

Sr(OH)2 + 2HCl --> 2H2O + SrCl2

hence two moles of HCl will react with one mole Sr(OH)2

So 0.0028 moles of HCl will react with 0.0028/2 moles of Sr(OH)2 = 0.0014 moles of Sr(OH)2

Hence moles of Sr(OH)2 after complete reaction = Initial moles - Moles reacted with HCl

Moles of Sr(OH)2 left = 0.006 - 0.0014 = 0.0046 moles

total volume = Volume of Hcl + volume of Sr(OH)2 = 28 + 60 = 88 mL

Concentration of Sr(OH)2 after reaction = moles / Volume in Litres = 0.0046 X 1000 / 88 = 0.052 molar

[OH-]= 2 X molarity of Sr(OH)2   {as each mole of this base will give two moles of hydroxide ions}

[OH-] = 2 X 0.052 = 0.104 molar

pOH = -log [OH-] = 0.98

pH = 14 - pOH = 14 - 0.98 = 13.12