When a small amount of acid is added to a non-buffe solution, there is a large change in pH. Calculate the pH when 26.9 mL of 0.0044 M HCl is added to 100.0 mL of pure water. Comment and hint in the general feedback.

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When a small amount of acid is added to a non-buffe solution, there is a large change in pH. Calculate the pH when 26.9 mL of 0.0044 M HCl is added to 100.0 mL of pure water. Comment and hint in the general feedback.

Answer 1

the correct answer is pH= 2.93

Answer 2

HCl is a strong acid.That means it separates. completely in the water as follows:hcl(aq)→h+(aq)+cl−(aq )When dissociation is complete, the amount of H+ produced is equal to the concentration of HCl.Therefore, we multiply the volume of HCl ( 26.9ml ) added by the concentration of HCl ( 0.0044 m ), and then divide it by the volume of water ( 100.0ml ).Note this factor.26.9ml/100ml is the dilution factor: [ HCl ] =..26.9ml/100ml x.0.0044 m =.1.18 x 10−3 mWe know that pH = -log [ H + ], in this case [ H + ] = [ HCl ] ( because HCl is a strong acid ), so we directly calculate the pH value of [ HCl ] as follows:pH = -log[H+ ] = -log[HCl ] = -log ( 1.18 x 10−3 M ) =. 2.92please note 2.92 is a lower phSo when we add strong acids to the water -- there is no buffer solution -- the pH suddenly drops.

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When a small amount of acid is added to a non-buffe solution, there is a large change in pH. Calculate the pH when 26.9 mL of 0.0044 M HCl is added to 100.0 mL of pure water. Comment and hint in the general feedback.

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