In: Chemistry
Calculate [OH?] and pH for strong base solution formed by mixing 15.0mL of 1.00
we know that
moles = molarity x volume (L)
moles of Ba(OH)2 = 1 x 10-2 x 15 x 10-3
moles of Ba(OH)2 = 1.5 x 10-4
Ba(OH)2 -----> Ba+2 + 2OH-
strong base so 100% dissociation
from the above reaction
moles of OH- = 2 x moles of Ba(OH)2
so
moles of OH- = 2 x 1.5 x 10-4
moles of OH- = 3 x 10-4
now consider NaOH
moles of NaOH = 7.4 x 10-3 x 43 x 10-3
moles of NaOH = 3.182 x 10-4
NaOH ----> Na+ + OH-
strong base so 100% dissociation
from the above reaction
moles of OH- = moles of NaOH = 3.182 x 10-4
now
total moles of OH-= moles of OH- from Ba(OH)2 + moles of OH- from NaOH
total moles of OH- = 3 x 10-4 + 3.182 x 10-4
total moles of OH- = 6.182 x 10-4
total volume = 15 + 43 = 58 ml
conc of [OH-] = total moles / total volume (L)
so
[OH-] = 6.182 x 10-4 / 58 x 10-3
[OH-] = 1.0658 x 10-2
so
[OH-] = 1.0658 x 10-2
pOH = -log [OH-]
pOH = -log 1.0658 x 10-2
pOH = 1.97
but
pH = 14 - pOH
so
pH = 14 - 1.97
pH = 12.03
so the pH is 12.03