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Calculate [OH?] and pH for strong base solution formed by mixing 15.0mL of 1.00

Calculate [OH?] and pH for strong base solution formed by mixing 15.0mL of 1.00

Solutions

Expert Solution

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

queen_honey_blossom answered 1 month ago
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