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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?

Respuesta :

Answer:

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

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