Question

If 0.00300 mol NaOH and 0.00300 mol Ba(OH)2 are completely dissolved in water to make1.00 L of solution at 25oC, what is the pOH of the solution?

Answer 1

first consider NaOH

NaOH ---> Na+ + OH-

NaOH is a strong base , so 100% dissociation

we can see that

moles of OH- = moles of NaoH = 0.003

now

consider Ba(OH)2

Ba(OH)2 ---> Ba+2 + 2OH-

Ba(OH)2 is also a strong base , so 100% dissociation
we can see that

moles of OH- = 2 x moles of Ba(OH)2

so

moles of OH- = 2 x 0.003

moles of OH- = 0.006

now

total moles of OH- = 0.003 + 0.006 = 0.009

now

volume = 1 L

we know that

conc = moles / volume (L)

so

[OH-] = 0.009 / 1 = 0.009

now

we know that

pOH = -log [OH-]

pOH = -log 0.009

pOH = 2.045757

so

pOH of the solution si 2.045757