To find the pH of the resulting solution when pH of 20 ml of 0.2 M HNO3 mixed with 40 ml of 0.1 M Ba(OH)2, find the number of moles of H+ and OH– for each given compound.
We know, Moles = Molarity x Volume
Since, Ba(OH)2 on dissociation give 2 OH– ion while HNO3 give 1 H+ ion, so
Moles of HNO3 = 0.2 M x 0.020 L = 0.004 moles
Moles of H+ ion = 0.004 moles
Moles of Ba(OH)2 = 0.1 M x 0.040 L = 0.004 moles
Moles of OH– ions = 2 x 0.004 moles = 0.008 moles
Now, 0.004 moles H+ ions are neutralized by 0.004 moles OH– ions.
Remaining OH– ions = 0.008–0.004 = 0.004 moles
Total Volume = 20 + 40 = 60 ml = 0.060 L
[OH–] = 0.004/0.060 M = 0.067 M
pOH = -log[OH–] = – log[0.067] = 1.17
pH = 14 – 1.17 = 12.83
Hence the pH of the resulting solution is 12.83
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