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What is the concentration of hydroxide ions after 50.0 mL of 0.250 M NaOH is added...

What is the concentration of hydroxide ions after 50.0 mL of 0.250 M NaOH is added to 120 mL of 0.200 M Na2SO4?

What is the concentration of hydroxide ions after 50.0 mL of 0.250 M NaOH is added to 120 mL of 0.200 M Sr(OH)2?

Solutions

Expert Solution

a) The moles of hydroxide ions added = Molarity of OH- X volume of OH- in litres = 0.250 X 50 X 10^-3 = 12.5 millimoles

The volume of solution = volume of OH- + volume of Na2SO4 = 50 + 120 = 170 mL

Now the new concentration of hydroxide ions = Initial moles of OH- / total volume

Concentration of hydroxide ions = 12.5 / 170 = 0.0735 Molar

b) The moles of hydroxide ions added = Molarity of OH- X volume of OH- in litres = 0.250 X 50 X 10^-3 = 12.5 millimoles

The moles of Sr(OH)2 added = Molarity of Sr(OH)2X volume of Sr(OH)2 in litres = 120 X 0.2 = 24 millimoles

Sr(OH)2 is a strong acid and will dissociates completely

1 mole of Sr(OH)2 will give two moles of Hydroxide ions

so 24 millimoles of Sr(OH)2 will give = 48 millimoles of OH-

total moles of OH- = 12.5 + 48 = 60.5 millimoles

total volume of solution = 120 + 50 = 170 mL

concentration of hydroxide ions = Moles / volume = 60.5 / 170 = 0.356 Molar

queen_honey_blossom answered 1 year ago
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