ENVE 512- INDUSTRIAL WATER and WASTEWATER TREATMENT

ENVE512- INDUSTRIAL WATER and WASTEWATER TREATMENT

NeutralizationHomework

Fall2015

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Figure1: Titration curve for reaction between 1N NaOH and 0.42N HF. Thedash shows the titration endpoint for the reaction while the dotshows the point of inflection.

Figure2: Titration curve for reaction between 1N NaOH and 0.18N HF. Thedash shows the titration endpoint for the reaction while the dotshows the point of inflection.

Forthe reaction using 0.42N HF, 10ml of NaOH had been used for theendpoint to be reached.

Thereforemoles of HF used at endpoint, (0.42M x 25ml)/ 1000ml = 0.0105moles

Molesof NaOH used at endpoint, (1M x 10ml)/ 1000ml = 0.01 moles

Hencethe mole ratio for this reaction is, (0.0105/ 0.01) = 1.05

Forthe reaction using 0.18N HF, 8.1 ml of NaOH had been used for theendpoint to be reached.

Thereforemoles of HF used at endpoint, (0.18M x 50ml)/ 1000ml = 0.009 moles

Molesof NaOH used at endpoint, (1M x 8.1ml)/ 1000ml = 0.0081moles

Hencethe mole ratio for this reaction is, (0.009/ 0.0081) =1.11

Thetheoretical value for the mole ratio between a strong base and astrong acid is 1.0 therefore, the experimental value is close.

Sincethe mole ratio is 1, for 6N NaOH to neutralize 0.6N one would require0.1liters of the base for every cubic meter (Kealey,2013).

References

Kealey,D. (2013).&nbspExperimentsin modern analytical chemistry.Springer.