What is the pH of 20 ml of 0.2 M HNO3 mixed with 40 ml of 0.1 M Ba(OH)2?

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

Read More: Find final concentration of this new solution.

Visit here