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Calculate the pH and pOH of the following strong base solutions: (a) 0.030 M NaOH, (b) 0.12 M Ba(OH)2, (c) 2.2 M NaOH

Manika gupta
one month ago
Dear Student
You can see your solution here

[a]
pOH = -log[OH- ] = -log(0.030 ) = 1.522
pH = 14.00 – pOH = 14.00 – 1.522 = 12.478

[b]
Barium hydroxide is a strong base for both stages of dissociation:

Ba(OH)2(s)→Ba2++2OH−

So the solution will have 0.2 M hydroxide ions. Now use the autodissociation product for water:

Ba(OH)2(aq) → Ba2+(aq) + 2 OH–(aq) (strong base)

0.045 M Ba(OH)2= 2(0.12) M OH–= 0.024 M OH–

pOH = –log[OH–] = –log[0.24] = 0.619

[c]
pOH = -log[OH- ] = -log(2.2 ) = -0.34
pH = 14.00 – pOH = 14.00 +0.34 = 14.34