Question

What is the pH of the solution when 25.5 mL 0.126 M Cu(NO3)2 is mixed with 10.0 mL 0.204 M NaOH? The Ksp of Cu(OH)2 is 1.1 x 10-15 at 25 degrees celsius.

Answer 1

Ans:

Millimoles of NaOH taken = Molarity x volume(mL) = 0.204 M x 10.0 mL = 2.04 mmol

Millimoles of Cu(NO₃)₂ taken = Molarity x Volume(mL) = 0.126 M x 25.5 mL = 3.213 mmol

Molar ratio: Cu(NO₃)₂ : NaOH = 3.213 : 2.04 = 1.575 : 1

Reaction: Cu(NO₃)₂ (aq) + 2 NaOH (aq) --> Cu(OH)₂ (s) + 2 NaNO₃ (aq)

For the completion of reaction we need Cu(NO₃)₂ : NaOH = 1:2.

But we have taken Cu(NO₃)₂ : NaOH = 1.575:1.

So, NaOH which is present in less amount is a limiting reagent.

During a reaction limiting reagent will be completely consumed.

So, 2.04 mmol of NaOH reacts with (2.04/2) = 1.02 mmol of Cu(NO₃)₂ to give 1.02 mmol of Cu(OH)_2.

Cu(OH)₂ being a sparingly soluble salt undergo partial dissociation in water.

If, ‘S’ M is the solubility of Cu(OH)₂ in water, then it gives ‘S’ M Cu²⁺ ions and ‘2S’ M OH^- ions in water.

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So, [OH^-] = 2S = 2 x 6.503 x 10^-6 M = 1.3006 x 10^-5 M

pOH = -log[OH^-] = -log(1.3006 x 10^-5) = 4.8859

pH = 14-pOH = 14-4.8859 = 9.1141