A careless laboratory technician wants to prepare 200.0 mL of a 0.025 m HCl (aq) solution but uses a volumetric flask of vol - ume 250.0 mL by mistake. (a) What would the pH of the desired solution have been? (b) What will be the actual pH of the solution as prepared?

Question

Jeremy Cross

Chemistry

12 March, 17:44

A careless laboratory technician wants to prepare 200.0 mL of a 0.025 m HCl (aq) solution but uses a volumetric flask of vol - ume 250.0 mL by mistake. (a) What would the pH of the desired solution have been? (b) What will be the actual pH of the solution as prepared?

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Answers (1)

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Jaydon · Answer 1

a) 0.9030 b) 1

Explanation:

PH is calculated by taken negative logarithms of hydrogen ion concentration of a solution; when HCl dissociate in one water, it will yield 1 mole of hydrogen ion concentration likewise if 0.025 m of HCl dissociates in water it will yield 0.025 mole of hydrogen concentration.

Molarity of HCl in the solution in 200 mL = mole / volume in litres = 0.025 / (200/1000) = 0.125 M

pH = logbase 10 (0.125^ - 1) = 0.903

Molarity of HCl in 250 mL = 0.025 / (250/1000) = 0.1 M

pH = logbase 10 (0.1^-1) = 1