Question

Consider the titration of 20.00 mL of 0.100 M HBr with 0.150 M KOH at 25 °C. What would be the pH of the solution when 20.00 mL of KOH have been added?

Answer 1

Moles of Acid or Base = Molarity x Volume in Ltr.

For Monoprotic acid HBr, Molarity = 0.1 M and Volume = 20.00 mL = 0.02 L

Moles of HBr = 0.1 x 0.02 = 0.002 moles.

For Monoacidic base KOH, Molarity = 0.15 M and Volume = 20.00 mL = 0.02 L

Moles of KOH = 0.15 x 0.02 = 0.003 moles.

Both HBr and KOH are strong electrolytes and hence ionizes completely.

Moles of H⁺ = Moles of HBr = 0.002 moles.

Moles of HO^- = Moles of KOH = 0.003 moles.

On mixing these two solutiions 0.002 moles of H⁺ are completely neutralized by 0.002 moles HO^- and excess of 0.001 moles HO^- will remain unneutralized.

Moles of free HO^- = 0.001 and Volume of solution = 0.02 L + 0.02 L = 0.04 L

Hence,

[HO^-] = # of Moles / Volume in L

[HO^-] = 0.001 moles / 0.04 L

[HO^-] = 0.025 moles/L

[HO^-] = 0.025 M.

By definition of pOH,

pOH = -log[HO^-]

pOH =- log(0.025)

pOH = 1.60

Then

pH + pOH = 14

pH + 1.60 = 14

pH = 14 - 1.60

pH = 12.40

pH of resultant solution is 12.40.

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